1000 ml = 1 L
20 ml = x L
cross multiply
x = 20/1000
x = 0.02 Litre

1000 mcg = 1 mg
250 mcg = x mg
cross multiply
x = 250/1000
x = 0.25 mg

1 kg = 2.2 lb
x kg = 160 lb
cross multiply
x = 160/2.2
x = 72.7 kg

The preperation requires that 500 mg of aspirin is to be contained in every 10 ml
500 mg = 10 ml
Total volume = 200 ml
X mg = 200 ml
cross multiply
x = 10,000 mg
recall 1 tablet = 500 mg
Y no of tablets = 10,000 mg
Y = 10,000/500
Y = 20 tablets.

First we convert 6 g to mg
6 * 1000 = 6,000 mg
30 capsules = 6000 mg
1 capsule = X mg
cross multiply
X = 200 mg

First we convert 5 g to mg
Recall 1000 mg = 1 g
X mg = 5 g
cross multiply
X = 5000 mg
Total quantity dispensed = 260 + 420 + 630 = 1310 mg
Quntity remaining = Total - quantity dispensed
Quantity remaining = 5000 - 1310
Quantity remaining = 3690 mg or 3.69 g

1 bottle = 95 mL
65 bottles = 95 * 65
6175 mL
divide by 1000 to express in Litre
6.175 mL.

Total weight = 20 g
First we convert 20 g to milligrams (mg)
1 g = 1000 mg
20 g = X mg
cross multiply
X = 20,000 mg
1 tablet = 250 mg
Y no of tablets = 20,000 mg
Y = 20,000/250
Y = 80 tablets.

First express 0.002 % w/v in mass concentration using the relationship 1 % w/v means 1 g of solute in 100 ml of solution
Therefore 0.002 % means 0.002 g in 100 ml
0.002 g = 100 ml
X g = 25 ml
cross multiply
X = 0.0005 g
Finally we coonvert this value to mg (because the question calls for such)
1 g = 1000 mg
0.0005 g = Y mg
0.0005 * 1000
Y = 0.5 mg

First we find the relationship between all the units in the options
This standard relationship is given by;
1 kg = 1000 g
1 g = 1000 mg
1 mg = 1000 mcg
Therefore it follows that; 1 kg = 1000 g = 1,000,000 mg = 1,000,000,000 ng
Next we use the relationship that corrsponds to ng
1 kg = 1,000,000,000 ng
X kg = 2000 ng
cross multiply
X = 2,000 / 1,000,000,000 kg
X = 0.000002 kg
Recall 1 kg = 1000 mg
0.000002 kg = Y mg
Y = 0.000002 * 1000
Y = 0.002 mg
So the options of 2 kg and 2 mg are not correct (given that our ansers were 0.000002 kg and 0.002 mg)
Recall 1 mg = 1000 mcg
0.002 mg = Z mcg
cross multiply
Z = 1000 * 0.002 mg
Z = 2 mcg (this is our final answer)

1 Mega unit = 1 * 10⁶ units
2 Mega units = 2 * 10⁶ units
25 ml = 2 * 10⁶ units
1 ml = X units
cross multiply
X = 2,000,000 / 25
X = 80,000 units
Answer 80,000 units per ml

0.05 mg of digoxin ---- 1 ml of mixture
X mg of digoxin ---- 4 ml of mixture
cross multiply
X = 4 * 0.05 mg
X = 0.2 mg
Convert mg to mcg
0.2 * 1000
200 mcg

First we convert 175 lb to kg
1 kg ---- 2.2046 lb
X kg ---- 175 lb
cross multiply
X = 175 / 2.2046
X = 79.5 kg
4 mg ---- 1 kg
Y mg ---- 79.5 kg
cross multiply
Y = 318 mg
318 mg every 4 hours
For 24 hours (6 times a day )
318 * 6 = 1909 mg or 1.9 g

Recall the expression
No of moles (n) = Mass in grams (M) / Molecular weight (W)
Next convert 50 mg to g
1 g = 1000 mg
X g = 50 mg
cross multiply
X = 50 / 1000
also written as 50 * 10^-3
n = 50 * 10^-3 / 318
n = 1.57 * 10^-4 moles

Recall the expression
No of moles (n) = Mass in grams (M) / Molecular weight (W)
Mass (M) = 206.3 * 10
Mass (M) = 2063 g

Recall the expression
(F - 32) / 9 = C / 5
(104 - 32) / 9 = C / 5
72 / 9 = C / 5
cross multiply
360 / 9
Temperature in degree centigrade = 40
for kelvin
Temperature in Kelvin = 273 + temperature in centigrade
Temperature in kelvin = 273 + 40 = 313 K

Recall percentage means a unit expressed in 100
1 ---- 2500
X ---- 100
cross multiply terms
X = (100 * 1) / 2500
X = 0.04
value in percentage = 0.04 %

Recall percentage weight in volume means 1 g in 100 ml
So if 250 mcg of solute ---- 1 ml of solution
X g of solute will be in ---- 100 ml of solution
Before cross multiplying terms recall that the weight given is in "mcg" not gram
Convert 250 mcg to g
250 * 10^-6
X = 250 * 10^-6 * 100
value in percentage = 0.025 % w/v

Recall percentage weight in volume means 1 g in 100 ml
So if 1 is in ---- 300
X g of solute will be in ---- 100 ml of solution
cross multiply terms
X = 100 / 300
value in percentage = 0.33 % w/v

Recall percentage weight in volume means 1 g in 100 ml
So if 1.5 g of solute ---- is dissolved 120 ml of solution
X g of solute will be in ---- 100 ml of solution
cross multiply terms
x = (100 * 1.5 ) / 120
X = 1.25 %

Recall percentage weight in volume means 1 g in 100 ml
2 Parts per million means 2 parts (g in this case) in 1,000,000 ml
So if 2 g of solute ---- is dissolved 1,000,000 ml of solution
X g of solute will be in ---- 100 ml of solution
cross multiply terms
X = (100 * 2 ) / 1,000,000
X = 2 * 10^-4 % w/v or 0.0002 % w/v

Recall percentage weight in volume means 1 g in 100 ml
So if 0.05 g of solute ---- is dissolved 100 ml of solution
X g of solute will be in ---- 1,000,000 ml of solution
cross multiply terms
X = (1,000,000 * 0.05 ) / 100
X = 500 ppm

Recall 0.5 % percentage weight in volume means 0.5 g in 100 ml
So if 0.5 g of solute ---- is dissolved 100 ml of solution
X g of solute will be need for ---- 25 ml of solution
cross multiply terms
X = (25 * 0.05 ) / 100
X = 0.125 g
Next we convert this value to milligrams as the question demands
Given that 1000 mg makes 1 gram
Y mg will be in .......... 0.125 g
cross multiply terms
Y = (0.125 * 1000) / 1
Y = 125 mg

We would apply the dilution formula C1V1 = C2V2
where C1, C2 = initial and final concentrations respectively
where V1, V2 = initial and final volumes respectively
C1 = 15 % v/v
C2 = ? (our unknown)
V1 = 500 ml
V2 = 1750 ml
500 * 15 = C2 * 1750
C2 = 7500/1750
C2 = 4.29 % v/v

We would apply the dilution formula C1V1 = C2V2
where C1, C2 = initial and final concentrations respectively
where V1, V2 = initial and final volumes respectively
C1 = 1:20
C2 = ? (our unknown)
V1 = 50 ml
V2 = 1000 ml
We would convert 1 in 20 to percentage strength so we can apply our formula
1 in 20 (w/v) means 1 g in 20 ml
1 ---- 20
X ---- 100 ml
X = (100 * 1 )/ 20
X = 5 % w/v
5 * 50 = C2 * 1000
C2 = 250/1000
C2 = 0.25 % w/v

We would apply the dilution formula C1V1 = C2V2
where C1, C2 = initial and final concentrations respectively
where V1, V2 = initial and final volumes respectively
C1 = 1:400
C2 = 1:2000
V1 = ? (our unknown)
V2 = 4 L (equivalent to 4000 ml)
We convert would 1 in 400 to percentage strength so we can apply our formula
1 in 400 (w/v) means 1 g in 400 ml
1 ---- 400
X ---- 100 ml
X = (100 * 1 )/ 400
X = 0.25 % w/v
We also convert 1 in 2000 to % w/v using the same method
This gives 0.05 % w/v
0.25 * V1 = 0.05 * 4000
V1 = 200/0.25
V1 = 800 ml

Given that a dilution is involved we would apply the dilution formula C1V1 = C2V2
where C1, C2 = initial and final concentrations respectively
where V1, V2 = initial and final volumes respectively
In this question we are looking for a concentration (C1) from which a volume V1 (10 ml)
could be taken and diluted to 1000 ml (V2) to give a new concentration C2 (0.15 % w/v)
C1 * 10 = 0.15 * 1000
C1 = 150/10
C1 = 15 % w/v
We have found the concentration of formaldehyde required, but the question requests for the volume of formalin BP required
To find that recall that formalin BP contains 36 % w/w of formaldehyde
Hence 36 g of formaldehyde is contained in 100 g of formalin
Note too that the concentration of formaldehyde is 15 % w/v
This means that 15 g of formalin is contained in 100 ml of formaldehyde solution
The preperation is for 100 ml (see first line of question) so we only need this 15 g of formaldehyde
But what mass of fomalin would be need to produce this mass of formaldehyde
Using 36 g ---- 100 g
15 g ---- X g
cross multiply
X = 41.67 g
In simple terms 41.67 g of formalin BP would be need to produce 15 g of formaldehyde
The question however requests for volume not mass of formalin BP required
Recall Density = mass / volume
Volume = mass/density
Volume = 41.67/1.08
Volume = 38.6 ml of formalin BP
Working backwards we can say 38.6 ml of formalin BP would be required to produce 15 g of formaldehyde, this 15 g of formaldehyde would subsequently be dissolved in 100 ml of a solvent, to produce a 15 % w/v concentration (a stock), from which 10 ml can be taken, diluted to 1000 ml in other to produce 0.15 % w/v solution.

We would apply the dilution formula C1V1 = C2V2
where C1, C2 = initial and final concentrations respectively
where V1, V2 = initial and final volumes respectively
C1 = 1 % w/v
C2 = ? (our unknown)
V1 = 10 ml
V2 = 1 L (equivalent to 1000 ml)
1 * 10 = C2 * 1000
C2 = 10 / 1000
C2 = 0.01 % w/v (first dilution)
0.01 * 10 = C2 * 1000
C2 = 0.1 / 1000 = 0.0001 % w/v (second dilution)
We next convert the final C2 to mg/l
Note 0.0001 % w/v means 0.0001 g in 100 ml
Next we convert 0.0001 g to mg by multiplying by 1000
0.0001 * 1000 = 0.1 mg
We now have 0.1 mg in 100 ml
The question demands the value in mg per Litre
If 0.1 mg is contained in ---- 100 ml
Y mg would be contained in ---- 1000 ml (1 L)
cross multiply
Y = (0.1 * 1000) / 100
Y = 1 mg / L

Recall percentage weight in volume means 1 g in 100 ml
0.6 Parts per million means 0.6 parts (g in this case) in 1,000,000 ml
So if 0.6 g of solute ---- is dissolved 1,000,000 ml of solution
X g of solute will be in ---- 100 ml of solution
cross multiply terms
X = (100 * 0.6 ) / 1,000,000
X = 6 * 10^-5 % w/v or 0.00006 % w/v

First we find the weight of 100 ml saturated potassium iodide
Using the relationship Density = mass / volume
Mass = volume * density
Mass of saturated potassium iodide solution = 100 * 1.7 = 170 g
So 100 ml of the product weighs 170 g but contains only 100 g of KI
The question demands for % w/w that means we must find a weight that would be contained in 100 g, what we currently have is a weight contained in 170 g
So if 100 g is contained in ---- 170 g of product
X g would be contained in ---- 100 g of product
cross multiply out terms
X = (100 * 100 )/170
X = 58.8 % (59 %)

We would work out each of the individual volume of ethanol
for 95 % = (95/100) * 300 = 285 ml
for 70 % = (70/100) * 1000 = 700 ml
for 50 % = (50/100) * 200 = 100 ml
Add up the corresponding alcohol volumes = 285 + 700 + 100 = 1085 ml
Add up the volumes of each solution = 300 + 1000 + 200 = 1500
Percentage v/v = (volume of dissolved liquid / total volume ) * 100
Percentage alcohol = (1085 / 1500) * 100 = 72 % v/v

0.001 % w/v means 0.001 g in 100 ml
0.001 g is in ---- 100 ml
X g would be in 50 ml
X = (50/100) * 0.001 = 0.0005 g
convert 0.0005 g to mg by multiplying by 1000
1000 * 0.0005 = 0.5 mg

Total weight of tablets = Number of tablets * individual weight
Total weight of tablet = 4 * 200 mg = 800 mg = equivalent to 0.8 g
Percentage concentration = (mass in gram / volume) * 100
Percentage concentration = (0.8 / 1000)* 100
Percentage concentration = 0.08 % w/v

We would work out each of the individual volume of ethanol
for 90 % = (90/100) * 400 = 360 ml
for 70 % = (70/100) * 1000 = 700 ml
for 50 % = (50/100) * 500 = 250 ml

Add up the corresponding alcohol volumes = 360 + 700 + 250 = 1310 ml
Add up the volumes of each solution = 400 + 1000 + 500 = 1900
Percentage v/v = (volume of dissolved liquid / total volume ) * 100
Percentage alcohol = (1310 / 1900) * 100 = 68.9 % v/v

First we find the total weight of 3 tablets
1 tablet = 200 mg
3 tablets = 3 * 200 = 600 mg
convert 600 mg to g by dividing by 1000
weight of 3 tablets = 0.6 g
Percentage concentration = mass in g / volume
Percentage concentration = (0.6 g / 1500 ) * 100
Percentage concentration = (0.6 / 1500) * 100 = 0.12 % w/v

First we determine the amount of water consumed by the animal in ml
1 L = 1000 ml
10 L = X ml
cross multiply
X = 10 * 1000 = 10,000 ml
0.35 ppb = 0.35 g in 1,000,000,000 ml
X g in ---- 10,000 ml
X g = (0.35 * 10,000) / 1,000,000,000 = 0.0000035 g
Equivalent to 0.0035 mg or 3.5 mcg

First we determine the mass of ephedrine in the total volume of preparation
0.25 % w/v means 0.25 g in 100 ml
X g = 30 ml
cross multiply
X = 30 / 100 * 0.25 = 0.075 g
The stock solution is stated to be 1 : 60
1 g in 60 ml
0.075 g in ---- X ml
cross multiply
X g = 0.075 * 60 = 4.5 ml

1.7 ppm means 1.7 g in ---- 1,000,000 ml
X g would be contained in ---- 50 ml vial
cross multiply
X = (50 * 1.7) / 1,000,000
X = 0.000085 g or 0.085 mg or 85 mcg

Recall 25 % v/v means 25 ml of active (methylsalicylate) in 100 ml of preperation
Since the preparation is exactly 100 ml it follows that the volume of active is also 25 ml
Next we find the weight of the "active"(methyl salicylate) contained in that 25 ml
Recall Density = mass / volume
1.18 = mass / 25
Weight of methylsalicylate in preparation = 29.5 g

Recall density = mass in g / volume
Since the question demands the volume of liquid paraffin required the density of the finished ointment is irrelevant
Volume = mass / density
Volume = 10 / 0.85
Volume = 11.76 ml (answer in ml because density was given in g/ml)

Recall density = mass in g / volume
Volume = mass / density
Volume = 20 / 0.910
Volume = 21.97 ml (answer in ml because density was given in g/ml)

Recall density = mass in g / volume
Volume = mass / density
Next we convert 3.5 kg to g
Recall 1 kg = 1000 g
3.5 kg = 3.5 * 1000 = 3500 g
Volume = 3500 / 0.922
Volume = 3796 ml (answer in ml because density was given in g/ml)

If 0.8 mg of prednisolone is contained in ---- each ml
X mg is contained in ---- 100 ml of preparation
cross multiply out terms
X = 0.8 * 100 = 80 mg
1 tablet contains ---- 5 mg of prednisolone
Y no of tablets is contained in ---- 80 mg
Y = 80 / 5 = 16 tablets

If 10,000 units is contained in ---- 1 ml of nystatin
X unit is contained in ---- 2000 ml of nystatin
cross multiply out terms
X = 20,000,000
1 mg of nystatin contains ---- 4,350 units of nystatin
Y mg of nystatin will contain ---- 20,000,000 units
cross multiply out terms
Y = 4,598 mg

Recall density = mass / volume
0.85 = mass / 97
cross multiply out terms
mass = 82.45 g

0.05 mg of digoxin is contained per ---- 1 ml
X mg will be contained in ---- 5 ml
cross multiply out terms
X = 5 * 0.05 = 0.25 mg
Convert this value to mcg by multiplying by 1000
X = 250 mcg

Recall 5 % w/w means 5 g of active in ---- 100 g of preparation
X g of glycerin would be contained in ---- 500 g of ointment
cross multiply terms
X = 25 g of glycerin
Recall density = mass in g / volume
1.25 = 25 / volume
cross multiply out terms
Volume = 25 / 1.25
Volume = 20 ml

Lets focus first on the preparation to be made 4 litres of a 10 % w/v of hydrochloric acid
Recall 10 % w/v means 10 g of hydrochloric acid in ---- 100 ml of preparation
X g would be contained in ---- 4,000 ml of preparation
cross multiply terms
X = 400 g of hydrochloric acid
so we require 400 g of hydrochloric acid in our final preparation
but we have an available stock of 36.8 % w/w concentrated hydrochloric acid
Recall 36.8 % w/w means 36.8 g in ---- 100 g of preparation
400 g of hydrochloric acid would be contained in ----- Y g of concentrated hydrochloric acid
cross multiply out terms
Y = 1087g
1087 g of concntrated HCL is required to produce 400 g of HCL need for a 10 % w/v dilute HCL solution
The question however demands the volume of the concentrated HCL needed, not its weight
Recall Density = mass / volume
Volume = 1087 / 1.19
Volume = 913.4 ml

First we find the weight of the solution
Recall Density = mass / volume
1.5 = mass / 100
cross multiply terms
mass = 150 g
Percentage w/w = weight of active / weight of preparation * 100
Percentage w/w = 120 / 150 * 100 = 80 % w/w

If 100 g of KI is contained in ---- 100 ml of solution
X g of KI would be contained in ---- 1000 ml of solution
cross multiply terms
mass = 1.7 * 1000
mass = 1700 g
Total weight of preparation = 1700 g
Weight of KI = 700 g
Therefore weight of water needed = Total weight - weight of active
Volume of water = 1700 - 700 = 1000 g
Given that density of water is 1 g/ml, volume of water needed = 1000 ml

First we find the weight of the solution
Recall Density = mass / volume
1.5 = mass / 100
cross multiply terms
mass = 150 g
Percentage w/w = weight of active / weight of preparation * 100
Percentage w/w = 120 / 150 * 100 = 80 % w/w

First we find the weight of acetic acid in the solution
Recall Density = mass / volume
1.04 = mass / 1000
cross multiply terms
Weight of solution = 104 g
Recall 33 % w/w acetic acid means 33 g of acetic acid in 100 g of acetic acid BP
If 33 g of acetic acid is in ---- 100 g of acetic acid BP
X g of acetic acid would be in ---- 104 g of acetic acid BP
cross multiply out terms
X = (104 * 33) / 100
X = 34.3 g of acetic acid

First we find the weight of glacial acetic acid in the solution
Recall Density = mass / volume
1.05 = mass / 1000
cross multiply terms
Weight of solution = 1050 g
Recall 99 % w/w glacial acid BP means 99 g of acetic acid in 100 g of glacial acetic acid BP
If g of acetic acid is in ---- 100 g of glacial acetic acid BP
X g of acetic acid would be in ---- 1050 g of acetic acid BP
cross multiply out terms
X = (1050 * 99) / 100
X = 1039.5 g of acetic acid

Recall Density = mass / volume
0.892 = 20 / Volume
cross multiply terms
Volume = 22.4 ml

Recall 5 % v/v means 5ml of propylene glycol in 100 ml of preparation
If 5 ml of PG is in ---- 100 ml of preparation
X ml of PG would be contained in ---- 50 ml of preparation
cross multiply terms
X = 2.5 ml
Hence 2.5 ml of PG is required, next we find the corresponding weight of PG required
Density = mass / volume
1.036 = mass / 2.5
cross multiply out terms
Mass = 2.59 g

Recall Density = mass in g / volume in cm³(same as milliitres (ml))
First we find the volume of the cylindrical tablet using
Volume = πr²h
Volume = 3.142 * 0.2² * 1.2
Volume = 0.1508 cm³
Mass = 180 mg or 0.18 g (Conversion by dividing by 1000)
Density = 0.18 / 0.1508
Density = 1.193 g/ml

Each 10 ml of mixture contains ---- 200 mg of allopurinol
200 ml of mixture will contain ---- X g of allopurinol
cross multiply terms
X = (200 * 200) / 10 = 4000 mg of allopurinol
1 tablet ---- 1OO mg of allopurinol
Y tablets ---- 4000 mg
cross multiply terms
Y = 40 tablets

First we must work out how much atropine sulphate is required to dispense 50 ml of the mixture
Each 5 ml contains -------- 60 mcg
50 ml of mixture will contain ---- X mcg of atropine sulphate
cross multiply terms
X = (50 * 60) / 5 = 600 mcg of atropine sulphate (equivalent to 0.6 mg)
1 ampoule contains 1.2 mg in every ---- 1 ml
0.6 mg will be contained in ----- Y ml
cross multiply terms
Y = 0.6 / 1.2 = 0.5 ml

Preservative in the mixture = 0.5 %
Recall 5 % w/v means 5g in 100 ml
5 g of preservative is in every ---- 100 ml of mixture
X g of preservative will be in ----- 200 ml of mixture
cross multiply terms
X = (200 * 0.5) / 100 = 1 g of preservative

CHB = MHP + PHB
CHB = 65 + 35 = 100 g
0.2 % CHB means 0.2 g of CHB in 100 g of cream
X g of CHB will be in ----- 50 g of cream
cross multiply terms
X = (0.2 * 50) / 100 = 0.1 g of CHB in cream
Next we find the amount of MHB in the cream using siimple proportion
Amount of MHB = Proportion of MHB / Proportion of CHB (MHB + PHB) * 0.1 (Proportion of actual CHB in 50 g of cream)
Amount of MHB = (65 / 100) * 0.1 = 0.065 g of MHB or 65 mg

Recall 0.6 % w/v means 0.6 g of atropine in 100 ml of atropine methonitrate alcoholic solution
The product however requires 0.3 mg of atropine in ---- 5 ml of mixture
X mg would be contained in ----- 100 ml of mixture
Cross multiply terms
X = (100 * 0.3) / 5 = 6 mg
Required weight of atropine = 6 mg
Next we convert this weight to g by dividing by 1000 = 0.006 g
If 0.6 g of atropine is in ---- 100 ml of areopine methonitrate solution
0.006 g (required) would be contained in ---- Y ml of atropine methonitrate solution
Cross multiply out terms
Y = (0.006 * 100) / 0.6 = 1 ml

1 suppositoy contains ---- 2.5 mg ergotamine
2500 suppositories will contain ---- X g of ergotamine
Cross multiply terms
X = 2.5 * 2500
X = 6250 mg or 6.25 g

100 capsules requires ---- 35 g of ephedrine
150,000 capsules will requiire ---- X g of ephedrine
Cross multiply terms
X = 150,000 * 35 / (100)
X = 52,500 g or 52.5 kg

Total number of parts = 1.5 + 1.5 + 3 = 6 parts
Using simple proportion
weight of starch required = number of parts of starch / total parts * 100
weight of starch required = (1.5 / 6) * 100 = 25 g

Total number of parts = 6 + 1 + 3 = 10 parts
Using simple proportion
weight of aspirin = number of parts of aspirin / total parts * 100
weight of aspirin = (6 / 10) * 1.25 = 0.75 kg or 750 g

Total number of parts = 12 + 1.5 + 4.5 = 18 parts
Using simple proportion
weight of talc required = number of parts of talc / total parts * 100
weight of talc required = (12 / 18) * 50 = 33.4 g

Total number of parts = 24 + 1 = 25 parts
Using simple proportion
weight of Beeswax required = number of parts of Beeswax / total parts * 100
Weight of Beeswax already given as = 550 g
The challange is to find the weight of ointment it can produce
550 = (1 / 25) * X
X = 25 * 550 = 13,750 g or 13.75 kg

We would begin by finding the mass of phenol in calamine lotion
Recall 5 % w/v means 5 g of phenol in 100 ml of calamine solution
Given that the pharmacist is producing 5 Litres (5000 ml) of calamine the mass of phenol needed would be X
0.5 g ---- 100 ml
X g in --- 5000 ml
Cross multiply terms
X = 5000 * 0.5 / (100) = 25 g of phenol
The pharmacist is required to include an extra 0.5 % phenol hence an additional 25 g of phenol
Total weight of phenol required = 25 + 25 = 50 g
The pharmacist can obtain the needed phenol from 80 % w/w liquefied phenol available
Recall 80 % w/w means 80 g of phenol in 100 g of liquefied phenol
80 g of phenol ---- in 100 g of LP
50 g of phenol (total required) ---- in X g of LP
cross multiply terms
X = 62.5 g of liquefied phenol (LP)
Given that LP exists as a liquid not a solid we must find the correspondng volume
Recall Density = mass / volume
volume = mass / density = 62.5 / 1.057
Volume of LP pharmacist require = 59.1 ml

10 ml of mixture contains ---- 300 mg of Paracetamol
200 ml of mixture (required) will require ----- X mg of paracetamol
cross multiply terms
X = 6000 mg of paracetamol
1 tablet contains ---- 300 mg of paracetamol drug
Y no of tablets will contain ---- 6000 mg
cross multiply
Y = 6000 / 300 = 20 talets

10 ml of mixture contains ---- 300 mg of Paracetamol
200 ml of mixture (required) will require ----- X mg of paracetamol
cross multiply terms
X = 6000 mg of paracetamol
1 tablet contains ---- 300 mg of paracetamol drug
Y no of tablets will contain ---- 6000 mg
cross multiply
Y = 6000 / 300 = 20 talets

5 mg of pholcodine is contained in ---- 5 ml
X mg of pholcodine is contained in ----- 500 ml of linctus
cross multiply terms
X = 500 mg of pholcodine

Preparation to be made should contain 25 mg of codeine in ---- 5 ml of codeine phosphate (CP) linctus
X mg of codeine is contained in ----- 100 ml of linctus
cross multiply terms
X = 500 mg of codeine
stock contains 20 mg of codeine in ---- 5 ml of CP stock
Y mg in ---- 62 ml of CP stock
cross multiply terms
Y = 248 mg
The pharmacist needs 500 mg (X value), the stock only provides 248 mg (Y value)
Extra CP neeed = X - Y = 500 - 248
Extra codeine phosphate needed would be = 252 mg

Percentage error = (sensitivity / weight) * 100
Percentage error = (12.5 / 400) * 100 = 3.125 %

Percentage error = (sensitivity / weight) * 100
5 = (11.5 / weight) * 100
cross multiply terms
Weight = 1150 / 5 = 230 mg

Percentage error = (sensitivity / weight) * 100
15 = ( sensitivity / 60) * 100
cross multiply terms
Weight = 900 / 100 = 9 mg

Percentage error = (sensitivity / weight) * 100
Percentage error = (5 / 25) * 100 = 20 %

Sensitivity means weight that can cause a 1 scale deflection
If 50 mg causes ---- 5 scale deflection
X mg will cause ---- 1 scale deflection
cross multiply terms
X = 50 / 5 = 10 mg
Sensitivity = 10 mg

Percentage error = (sensitivity / weight) * 100
Percentage error = (12.5 / 500) * 100 = 2.5 %

Percentage error = (sensitivity / weight) * 100
5 = (15 / weight) * 100
cross multiply terms
Weight = 1500 / 5 = 300 mg

Percentage error = (sensitivity / weight) * 100
Percentage error = (10 / 50) * 100 = 20 %

Percentage error = (sensitivity / weight) * 100
Percentage error = (12.5 / 250) * 100 = 5 %

First we convert 5 g to mg by multiplying by 1000 = 5000 mg
Percentage error = (sensitivity / weight) * 100
Percentage error = (10 / 5000) * 100 = 0.2 %

Recall 1 in 50,000 means 1 g of methylene blue in 50,000 ml of solution
1 g is contained in ---- 50,000 ml solution
X g would be contained in ---- 400 ml of solution
cross multiply terms
X = 400 / 50,000
X = 0.0096 g = 9.6 mg

Notice that when diluting the mass does not change but the volume does
So if 50 mg is contained in ---- 10 ml
50 mg would also be contained in ---- 1000 ml of solution
concentration = 50 mg/1000 ml
Next we express this concentration in mcg/ml
50 mg is contained in ---- 1000 ml
X mg would be contained in ---- 1 ml
cross multiply
X = 0.05 mg
Finally we convert 0.05 mg to micrograms by multiplying by 1000 = 50 mcg
Concentration in mcg/ml = 50 mcg/ml

Recall 5 % w/v means 5 g of aluminium acetate in 100 ml of solution
So if 5 g of aluminium acetate is contained in ---- 100 ml of solution
X g would be contained in ---- 240 ml of solution
cross multiply terms
X = 12 g
Therefore the mass of aluminium acetate in the final solution = 12 g
Next we find the total volume
Total volume = Initial volume + dilution volume
Total volume = 240 ml + 2 Litres (2000 ml) = 2240 ml
Finally we find the new concentration using
Percentage concentration = mass in g / volume * (100)
Percentage concentration = 12 /2240 * (100) = 0.54 % w/v

The dilution formula can be applied for all questions involving dilution
The pharmacist is required to produce 1 L of NaCL solution of an unknown strength; our C1
When 20 ml of this solution of unknown strength is taken; our V1
And diluted to 100 ml; our V2
The resulting concentration is 0.9 % w/v; our C2
C1 * V1 = C2 * V2
C1 * 20 = 0.9 * 100
C1 = 4.5 %
The concentration the pharmacist requires = 0.45 % w/v
The pharmacist however needs to produce 1 L (1000 ml) of this stock solution
Recall 0.45 % means 0.45 g of NaCL in 100 ml of preparation
If 0.45 g is contained in ---- 100 ml of preparation
X g would be contained in ---- 1000 ml
cross multiply terms
X = 45 g (total weight of NaCL the pharmacist requires for the 1 L preparation)
The pharmacist needs to use the available NaCL tablets
1 tablet = 2.25 g
Y no of tablets = 45 g
cross multiply terms
Y = 45 / 2.25 = 20 tablets

We would apply the dilution formlula C1 * V1 = C2 * V2
C1 = 4 %
V1 = ? (our unknown)
C2 = 10 mg per ml (we can convert this value to % concentration giving 1 % w/v; see section 2)
4 * V1 = 1 * 100
V1 = 100 / 4 = 25 ml

Notice in the question we are working with both sodium nitrite (NaNO₂) and nitrite ion (NO₂)
Sodium nitrite produces nitrte ion as shown below

NaNO₂ NO₂ + Na
Again notice that 0.1 % w/v NO₂ ion solution resulted from an unknown concentration of NaNO₂ solution
Lets find the concentration of the unknown NaNO₂ that gave off 0.1 % w/v NO₂
To do this we would utilize their molecular weights as equivalent proportions
We must however first find the molecular weight of NO₂
Molecular weight of NO₂ = molecular weight of NaNO₂ - molecular weight of Na = 69 - 23 = 46 g
Therefore using molecular proportions we can say;
69 g of NaNO₂ produced 46 g of NO₂
X % w/v of NaNO₂ will produce 0.1 % w/v solution of NO₂
cross multiply terms
X = (69 * 0.1 ) / 46 = 0.15 % w/v
Therefore we could say; 0.15 % NaNO₂ solution would be produced when 1 ml of an unknown concentration of sodium nitrite (NaNO₂) is diluted to 20 ml
Working backwards we could find the unknown concentration that produces 0.15 % NaNO₂
Lets apply the dilution formula C1 * V1 = C2 * V2
Where C1 = Unknown concentration
V1 = 1 ml
V2 = 20 ml
C2 = 0.15 % w/v
C1 * 1 = 0.15 * 20
C1 = 3 % w/v
The question however requires the mass needed to produce 1 Litre of this solution
Recall 3 % w/v means 3 g in 100 ml
If 3 g is contained in ---- 100 ml
Y g would be contained in ---- 1000 ml
cross multiply terms
Y = 30 g

We can conclude therefore that when 30 g of sodium nitrite (NaNO₂) is dissolved in 1 litre of water, the concentration is 3 % w/v.
When 1 ml of this solution is taken and diluted to 20 ml the resulting solution is 0.15 % w/v sodium nitrite solution, which contains 0.10 % w/v nitrite ion (NO₂)

We would apply the dilution formlula C1 * V1 = C2 * V2
C1 = 5 %
V1 = 120 ml
C2 = ? (our unknown)
V2 = 1000 + 120 = 1120
5 * 120 = C2 * 1120
C2 = 0.54 %

Alternatively we could say 5 % = 5 g in 100 ml
X g would be in 120 ml
X = 6 g
Percentage concentration = mass in g / volume * (100)
Percentage concentration = 6 / 1120 * (100)
Percentage concentration = 0.54 % w/v

We would apply the dilution formlula C1 * V1 = C2 * V2
C1 = 70 %
V1 = ? (unknown)
C2 = 30 %
V2 = 100
70 * V1 = 30 * 100
V2 = 3000 / 70 = 42.875 g
Next we convert this weight volume
Recall Density = mass / volume
Volume = mass / density = 42.857 / 1.290 = 33.2 ml

Think of this as a reverse calculation
We would find the concentration (C1) from which the stated dilution could be made
Next we must convert 1 in 2000 to percentage concentration (see section 2)
C1 * V1 = C2 * V2
C1 * 10 = 0.05 * 500
C1 = 2.5 % w/v
Next we find the weight 100 ml of solution would require
Recall 2.5 % w/v means 2.5 g in 100 ml
Given that the preparation calls for 100 ml we only need 2.5 g

50 mg per 10 ml
Note that diluting to 500 ml does not change the weight only the volume is affected
Mass concentration = mass / volume
Mass concentration = 50 / 500 = 0.1 mg per ml
Converting 0.1 mg to mcg multiply by 1000 (see section 1)
Mass concentration = 100 mcg/ml

Recall 1 in 25,000 means 1 g in 25,000 ml solution
If 1 g is contained in 25,000 ml solution
X g would be contained in 120 ml solution
Cross multiply terms
X = 0.0048 g or 4.8 mg

First we find the total volume of concentrated chloroform water our patient woud be taking
15 ml QDS means 15 ml 4 times daily
Total volume of concentrated chloroform water patient is receiving = 15 * 4 = 60 ml
Next we find the volume of chloroform water contained in the patients consumption volume of 60 ml mixture
Recall 2.5 % v/v means 2.5 ml of chloroform water in 100 ml of concentrated chloroform water
If 2.5 ml of chloroform water is contained in 100 ml chloroform water solution
X ml of chloroform water would be contained in 60 ml chloroform water solution
Cross multiply terms
X = 1.5 ml
Finally we find the exact amount of chloroform the patient would be taking based on the BP standards
Recall 10 % v/v means 10 ml of chloroform in 100 ml concentrated chloroform water
10 ml of chloroform is in 100 ml concentrated chloroform water
Y ml of chloroform would be contained in 1.5 ml
cross multiply terms
Y = 0.15 ml

Recall 10 % w/v means 10 g in 100 ml solution
If 10 g of HCL is contained in 100 ml HCL solution
X g of HCL would be contained in 4 Litres(4,000 ml) solution
Cross multiply terms
X = 400 g
From the available stength 36.8 g HCL is contained in 100 g HCL stock
400 g(needed) would be contained in X g of stock HCL
cross multiply terms X = 1087 g
Finally we find the required volume using
Volume = mass / density = 1087 / 1.19 = 913.4 ml

Recall 1 in 80 means 1 g in 80 ml solution
If 1 g is contained in 80 ml solution
X g would be contained in 100 ml solution
Cross multiply terms
X = 1.25 g in 100 ml equivalent to 1.25 % w/v

Recall 1 in 500 means 1 g in 500 ml solution
If 1 g is contained in 500 ml solution
X g would be contained in 1 ml solution
Cross multiply terms
X = 0.002 g in 1 ml
convert 0.002 g to mg by multiplying by 1000 = 2 mg
Answer in mg/ml = 2 mg/ml

We can solve using both alligation and dilution method

Alligation method
The rule for aligation method is as follows
The required concentration must lie between the maximum and minimum concentrations
Next we set up the alligation grid shown below

A		C
	X
B		D

Where A = Maximum concentration
Where B = Minimum concentration
Where X = Required concentration
Where C = Required - minimum concentration ( X- B)
Where D = Maxiumum - required concentration (A - X )

Lets input into table
Max = 70
Min = 0 (because we are diluting with water which has 0 % alcohol)
X = 45

70		45
	45
0		25

The values of C and D corresponds to the concentrations of A and B respectively.
The values of C and D are proportions, which could be expressed as parts, ml or grams depending on the context
The total proportion here would be = C + D = 45 + 25 = 70
Given that we are dealing with volumes, we would represent our proportions as ml from here on

Therefore we could say 45 ml (of 70 %) is required to prepare 70 ml (total volume of alcohol and water)
500 ml (of 70 %) will be required to produce X ml (total volume of alcohol and water)
cross multiply terms
X = 777.8 ml
Therefore in preparing 500 ml of 70 % v/v alcohol requires a total volume of 777.8 ml (alcohol + water) to reduce its concentration to 45 % v/v
Therefore volume of water required = Total volume - Volume of 70 % ethanol = 777.8 - 500 = 277.8 ml

Dilution method
C1 * V1 = C2 * V2
C1,V1 = 70 %, 500 ml
C2 = 45 %
V2 = ?
70 * 500 = 45 * V2
V2 = 777.8 ml
Volume of water needed = V2 - V1 = 777.8 - 500 = 277.8 ml

Alligation method see 6.16

8		5
	5
0		3

The total proportion = 5 + 3 = 8
Given that we are dealing with volumes, we would represent our proportions as ml from here on

Therefore we could say 5 ml (of 8 %) is required to prepare 8 ml total volume (containing benzoic acid and propylene glycol)
X ml (of 8 %) will be required to produce 100 ml of benzoic acid and propylene glycol mixture
cross multiply terms
X = 62.5 ml
Therefore in preparing 100 ml of 5 % benzoic acid, 62.5 ml of 8 % w/v benzoic acid should be used
Going further volume of pure propylene glycol required = Total volume - Volume of 8 % benzoic acid = 100 - 62.5 = 37.5 ml

Try this using the dilution method

Using the dilution method
C1 * V1 = C2 * V2
C1 = 90 % , C2 = 50 %
V1 = ? (unknown), V2 = 500 ml
90 * V1 = 50 * 500
V1 = 278 ml
Try this using alligation

Using the dilution method
C1 * V1 = C2 * V2
C1 = 70 % , C2 = 45 %
V2 = ? (unknown), V1 = 400 ml
70 * 400 = 45 * V2
V2 = 622.2 ml
Volume of water required = V2 - V1 = 622.2 - 400 = 222.2 ml
Try this using alligation

We are required to make a solution that will contain 1 ppm of F^-
To do this we would utilise drinking water to dilute a certain concentration
Given that drinking water already ccontains 0.5 ppm of flouride, our final solution only needs 0.5 ppm of flouride
1 - 0.5 = 0.5 ppm

Next we must find the concentration of flouride solution which can on dilution give our 0.5 ppm of flouride solution
We would call this unknown Concentration C1
C2 = 0.5 ppm
V1 = 30 ml
V2 = 1 L (1000 ml)
Applying the dilution formula C1 * V1 = C2 * V2
C1 * 30 = 0.5 * 1000
C2 = 16.67 ppm
Next we convert this value to gram per 200 ml (because our final volume of preparation is 200 ml)
Recall 16.67 ppm means 16.67 g in 1,000,000 ml
X g would be in 200 ml
X = 0.003334 g
Next we set up the molecular relationship between NaF and F because the question requires the amount of NaF needed not F
NaF = Na⁺ + F^-
42 g of NaF will produce 19 g of F^-
X g of NaF will be required to produce 0.003334 g of F^-
cross multiply terms
X = 0.00736 g
Multiplying by 1000 we get 7.4 mg

Recall 15 % w/v means 15 g of drug in 100 ml of solution
If 15 g is contained in 100 ml drug solution
X g would be contained in 250 ml drug solution
Cross multiply terms
X = 37.5 g of drug
Next we find the mass of the drug in alcohol
Mass = Volume * density = 250 * 0.952 = 238 g
Next we find the mass of alcohol using
Mass of alcohol = mass of alcohol in drug - mass of drug = 238 - 37.5 = 200.5 g
Finally we find the volume of alcohol using
Volume = mass / density = 200.5 / 0.798 = 251.25 ml

Recall 85 % w/w means 85 g of alcohol in 100 g
First we find the volume occupied by 85 g of pure ethanol
Volume = mass / density = 85 / 0.7937 = 107.09 ml
Next we find the volume occupied by 100 g of 85 % w/w
Volume = mass / density = 100 / 0.8337 = 119.35 ml
Percentage v/v ethanol = volume of 85 g / volume of 100 g * (100) = 107.09 / 119.95 * 100 = 89.2 % v/v

The solution of KI above can be expressed as 800 g/800 ml or 1 g/ml
If 1 g of KI is contained in 1 ml KI solution
X g of KI would be contained in 150 ml KI solution
Cross multiply terms
X = 150 g of KI per ml
Next we find the mass of the 150 ml solution using
Mass = volume * density = 150 * 1.7 = 255 g
Mass of water = Mass of 150 g solution - mass of KI = 255 - 150 = 105 g
Finally we convert this value to volume using
Volume = mass * density
Density of water = 1 g/ml
Volume of purified water required = 1 * 105 = 105 ml

Serial dilution detected we would apply the dilution formula
500 mg / 10000 mg * 1 g / 15 g * 1 g / 15 g * 1 g / 15 g * 500 mg = 0.0074 mg (answer in mg because all units canceals out except the last unit)
Converting to mcg by multiplying by 1000 = 1000 * 0.0074 mg = 7.4 mcg

Alligation method see 6.16

10		5
	5
0		5

Note that the minimum concentration in this case is 0 because we are mixing 10 % w/w benzoic acid with propylene glycol (0 % w/w benzoic acid)
The total proportion = 5 + 5 = 10
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 5 g (of 10 %) is required to prepare 10 g total weight (containing benzoic acid and propylene glycol)
X g (of 10 %) will be required to prepare 150 g of benzoic acid and propylene glycol mixture
cross multiply terms
X = 75 g
Therefore in preparing 100 ml of 5 % benzoic acid, 75 g of 10 % w/v benzoic acid should be used
Going further volume of pure propylene glycol required = Total volume - Volume of 10 % benzoic acid = 150 - 75 g = 75 g

Serial dilution detected we would apply the dilution formula
500 mg / 10000 mg * 1 g / 10 g * 1 g / 12 g * 300 mg = 0.125 mg (answer in mg because all units canceals out except the last unit in mg)
Converting to mcg by multiplying by 1000 = 1000 * 0.125 mg = 125 mcg

Serial dilution detected we would apply the dilution formula
800 mg / 10,000 mg * 1000 mg / 8000 mg * 1000 mg / 8000 mg * 1000 mg / 10,000 mg * 400 mg = 0.05 mg
Converting to mcg by multiplying by 1000 = 1000 * 0.05 mg = 50 mcg

Recall 1.25 % w/w means 1.25 g of sulphur in 100 g of sulphur ointment
If 1.25 g of sulphur is contained in 100 g of sulphur ointment
5 g of sulphur would be contained in X g of sulphur ointment
cross multiply terms
X = 400 g

1 powder contains 1 mg of prednisolone
125 powders X mg of prednisolone
cross multiply terms
X = 125 mg
1 prednisolone tablet contains 5 mg of prednisolone
Y no of tablets will contain 125 mg of prednisolone
cross multiply terms
Y = 25 tablets

First we work out how much prednisolone is required to make 100 powders
1 powder contains 1 mg of prednisolone
100 powders X mg of prednisolone
cross multiply terms
X = 100 mg

Next we work out how many tablets are required to produce 100 mg
1 prednisolone tablet contains 5 mg of prednisolone
Y no of tablets will contain 100 mg of prednisolone
cross multiply terms
Y = 20 tablets

Thirdly we work out the weight 100 powders
1 powder weighs 300 mg
100 powders will weigh Z mg
cross multiply terms
Z = 30,000 mg

Finally we determine the weight of 20 tablets
1 tablet weighs 150 mg
20 tablets will weigh W mg
cross multiply terms
W = 3000 mg

Weight of lactose required for 100 powders = weight of 100 powders - weight of prednisolone in 100 powders = Z - W = 30,000 - 3000 = 27,000 mg

Recall 12 % w/w means 12 g of active in 100 g of ointment
12 g of active is contained in 100 g of ointment
X g of active would be contained in 50 g of ointment
cross multiply terms
X = 6 g of active

Expressed in percentage we could say
6 g of active is contained in 75 g of ointment
Y g of active is contained in 100 g of ointment
cross multiply terms
Y = 8 g of active per 100 g of ointment equivalent to 8 % w/w

We could apply the dilution formula
C1 * M1 = C2 * M2 = 12 * 50 = C2 * 75
C2 = 600 / 75 = 8 % w/w

We would apply the dilution formula
0.3 / 10 * 1 / 10 * 1 / 12 * 1 / 12 * 0.3
6.24 * 10^-6 g = 0.00624 mg = 6.24 mcg

We could appy the dilution formula M1 * V1 = M2 * V2
Note 1:100 dilution is the same as 1 % w/w (see section 2)
Next we would convert the strength of the final mixture to percentage w/w
1 mg (0.001 g) is in 10 g of mixture
X g will be contained in 100 g of mixture
cross multiply terms
X = 0.01 g in 100 g of mixture (same as 0.01 % w/w)
Applying the dilution formula
1 * 10 = 0.01 * M2
M2 = 1000 g
Weight of diluent required = Weight after dilution - Weight before dilution = M2 - M1
Weight of diluent = 1000 - 10 = 990 g

First we find the dose required
2.5 mg per kg
X mg 15 kg
cross multiply terms
X = 37.5 mg

Next we find the total daily dose
37.5 mg per dose
37.5 * 2 = 75 mg per day
75 * 10 = 750 mg for 10 days
1 tablet contains 150 mg of roxithromycin
Y no of tablets will contain 750 mg of roxithromycin
cross multiply terms
Y = 5 tablets

1 wrapped powder contains 50 mg of active
20 wrapped powders will contain X mg of active
cross multiply terms
X = 1000 mg
1 wrapped powder weighs 500 mg
20 wrapped powder would weigh Y mg
cross multiply terms
Y = 10,000 mg
Weight of diluent = Total weight of powder - Total weight of active = Y - X = 10,000 - 1000
Weight of diulent = 9000 mg (equivalent to 9 g)

Recall 0.3 % w/v means 0.3 g of chlorine in 100 ml of calcium Hypochlorite solution
0.3 g is contained in 100 ml of calcium hypochlorite solution
X g of chlorine would be contained in 45 ml of calcium hypochlorite solution
cross multiply terms
X = 0.135 g
Note this is the amount of chlorine in calcium hpochlorite solution, it is also the amount of chlorine in the entire preparation
Next we find the total weight of solution
Total weight of soltion = 45 + 45 + 10 = 100 g

0.135 g of chlorine is contained in 100 g of solution
Y g of chlorine would be contained in 1 g of solution
cross multiply terms
Y = 0.00135 g (equivalent to 1.35 mg)

Using the BP standard we could say
20 g of chlorhexidine gluconate is in 100 ml of chlorhexidine gluconate solution
X ml of chlorhexidine gluconate would be contained in 5 ml of chlorhexidine gluconate solution
cross multiply terms
X = 1 g of chlorhexidine gluconate
Note this 1 g of chlorhexidine gluconate is the amount in 5 ml, and 5 ml of chlorhexidine gluconate solution is the volume in 100 g of the cream
Therefore 1 g of chlorhexidine gluconate is the amount in 100 g of cream
1 g in 100 g is the same as saying 1 % w/w (see section 2)
Answer in % w/w = 1 % w/w

Alligation method see 6.16

20		5
	5
0		15

Note that the minimum concentration in this case is 0 because we are mixing 20 % w/w salicylic acid with WSP (which contains 0 % w/w salicylic acid)
The total proportion = 15 + 5 = 20
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 5 g (of 20 %) salicylic acid is required to prepare 20 g total weight (containing salicylic acid and WSP)
X g (of 20 %) salicylic acid will be required to prepare 100 g of salicylic acid and WSP mixture
cross multiply terms
X = 25 g
Therefore in preparing 100 g of 20 % salicylic acid, 25 g of 20 % w/v salicylic acid should be used
Going further
Volume of pure WSP required = Total weight - weight of 20 % salicylic acid = 100 - 25 g = 75 g

Alligation method see 6.16

100		6
	10
4		90

Note the question doesnt say 64 g is the total weight of the ointment rather it says 64 g is the weight of the 4 % already available
Note also that the maximum concentration in this case is 100 because we can assume that the urea available for the preparation is 100 % pure

We can assign variables for the unknowns
Let the amunt of urea needed from 100 % = X
Let the total weight of ointment = Weight from 100 % (X) + Weight of 4 % = X + 64
Note finally that 64 g mentioned above is the total weight of 4 % urea available but not the weight of urea provided by 4 % urea compound

Infact we can calculate the weight of urea in 4 % urea compound using simple proportion
Recall 4 % means 4 g of urea in 100 g of urea compound
4 g of urea is contained in 100 g
Z g of urea would be contained in 64 g of urea compound
cross multiply
Z = 2.56 g of urea in 64 g urea compound (keep this for later)

The total proportion = 90 parts (of 4 %) + 6 parts (of 100 %) = 96 parts
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 6 g (of 100 %) urea is required to prepare 96 g total weight (containing 4 % and 100 % urea)
X g (of 100 %) urea will be required to prepare (64 + X) g total weight of 4 % and 100 % urea
cross multiply terms
96X = 6(64 + X)
96X = 384 + 6X
90X = 384
X = 4.27 g

Total weight of ointment = Weight of urea from 100 % (X) + Weight of 4 % = X + 64 = 4.27 + 64 = 68.27
Therefore in preparing 68.27 g of 10 % urea, 4.27 g of 100 % urea should be mixed with 64 g of 4 % urea
Let's recheck
10 % means 10 g of urea in 100 g of urea compound
or Total weight of urea / Total weight of compound * 100
(4.27 + 2.56) / (64 + 4.27) * 100 = 10 %

Serial dilution detected we would apply the dilution formula
0.6 g / 10 g * 1 g / 10 mg * 1 g / 12 g * 1 g / 6 g * 600 mg = 0.05 mg
Converting to mcg by multiplying by 1000 = 1000 * 0.05 mg = 50 mcg

Alligation method see 6.16

100		14
	20
6		80

Note that the maxmum concentration in this case is 100 because we are mixing 6 % w/w icthamol in WSP with 100 % pure icthamol
The total proportion by parts = 14 parts of 100 % pure icthamol + 80 % icthamol in WSP = 94
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 14 g (of 100 %) icthamol is required to prepare 94 g total weight (containing 100 % pure icthamol and 6 % icthamol in WSP)
X g (of 100 %) pure icthamol will be required to prepare 500 g of icthamol and WSP mixture
cross multiply terms
X = 74.5 g
Therefore in preparing 500 g of 20 % icthamol, 74.5 g of 100 % w/w icthamol should be used
Going further
Volume of 6 % icthamol needed = Total weight - weight of 100 % pure icthamol = 500 - 74.5 g = 425.5 g

Alligation method see 6.16

5		2
	2
0		3

Note that the minimum concentration in this case is 0 because we are mixing 5 % w/w hydrocortisone ointment with WSP (which contains hydrocortisone)
The total proportion = 2 + 3 = 5
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 2 g (of 5 %) hydrocortisone ointment is required to prepare 5 g total weight (containing hydrocortisone and WSP)
X g (of 5 %) hydrocortisone will be required to prepare 1000 g of salicylic acid and WSP mixture
cross multiply terms
X = 400 g
Therefore in preparing 1000 g of 2 % salicylic acid, 400 g of 5 % w/w hydrocortisone should be used
Going further
Volume of pure WSP required = Total weight - weight of hydrocortisone = 1000 - 400 g = 600 g

Alligation method see 6.16

0.928		0.05
	0.860
0.810		0.068

The total proportion = 0.05 + 0.068 = 0.118
Given that we are dealing with volumes, we would represent our proportions as ml from here on

Therefore we could say 0.05 ml (of 0.928 g/ml) mixture A is required to prepare 0.118 ml total volume (containing mixture A and B)
X ml (of mixture A) will be required to prepare 1000 ml of mixture A and B
cross multiply terms
X = 423.7 ml
Therefore in preparing 1000 ml of ethanol with a density of 0.860, 423.7 ml of mixture A should be used
Going further
Volume of mixture B required = Total volume - volume of mixture A = 1000 - 423.7 = 576.27 ml

Alligation method see 6.16

15		9.2
	13
3.8		2

The total proportion = 9.2 parts + 2 parts = 11.2 parts
Given that we are dealing with weights, we would express our proportions as g from here on

Therefore we could say 9.2 g of polysorbate 80 % is required to prepare 11.2 g of the emulsion
X g of polysorbate 80 will be required to prepare 150 g of the emulsion
cross multiply terms
X = 123.2 g of polysorbate 80

Going further
Required part of monosterate = Total weight - proportion of polysorbate 80 = 150 - 123.2 = 26.8 g

Alligation method see 6.16

20		5
	5
0		15

Note that the minimum concentration in this case is 0 because we are mixing 20 % w/w salicylic acid with WSP (which contains 0 % w/w salicylic acid)
The total proportion = 15 + 5 = 20
Given that we are dealing with weights, we would represent our proportions as g from here on

Therefore we could say 5 g (of 20 %) salicylic acid is required to prepare 20 g total weight (containing salicylic acid and WSP)
X g (of 20 %) salicylic acid will be required to prepare 100 g of salicylic acid and WSP mixture
cross multiply terms
X = 25 g
Therefore in preparing 100 g of 20 % salicylic acid, 25 g of 20 % w/v salicylic acid should be used
Going further
Volume of pure WSP required = Total weight - weight of 20 % salicylic acid = 100 - 25 g = 75 g

Alligation method see 6.16

11.6		7.4
	10.8
3.4		0.8

The total proportion = 7.4 + 0.8 = 8.2
Using simple proportion we can say
Percentage of PGM = Proportion of PGM / Total proportion * 8 = 7.4 / 8.2 * 8 % = 7.22
Percentage of PGM400 = Proportion of PGM400 / Total proportion * 8 = 0.8 / 8.2 * 8 = 0.78
Of the 8 % surfactant required in the emulsion 7.22 % should be PGM and 0.78 % should be PGM400

Alligation method see 6.16

1.00		0.375
	0.50
0.125		0.50

The total proportion = 0.375 + 0.50 = 0.875
Using simple proportion we can say
Amount of 0.125 % w/w iodine needed = Proportion of 0.125 % w/w iodine / Total proportion * Total weight = 0.5 / 0.875 * 60
Weight of 0.125 % required = 34.3 g

The other method in 7.18 would also give the same results

Alligation method see 6.16

1.5		0.5
	0.75
0.25		0.75

The total proportion = 0.5 + 0.75 = 1.25
Using simple proportion we can say
Amount of 0.25 % w/w iodine required = Proportion of 0.125 % w/w iodine / Total proportion * Total weight of preparation = 0.75 / 1.25 * 200 = 120 g
Amount of 1.5 % w/w iodine required = Total weight - weight of 0.25 % w/w iodine = 200 - 120 = 80 g

Try this: Find the weight of 1.5 % w/w iodine required using the proprtion method

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
Converting 500 mg to g we have 0.5 g
Applying the formula
Weight of base = (1.2 - (0.5 / 1.75)) * 1.2 * 1
Weight of base for 1 suppository = 1.097
Weight of base for 10 suppositories = 1.097 * 10 = 10.97 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would assume the base density to be = 1
We would convert 500 mg to g = 0.5 g
Applying the formula
Weight of base = (1.2 - (0.5 / 1.75)) * 1
Weight of base for 1 suppository = 0.9142
Weight of base for 15 suppositories = 0.9142 * 15 = 13.71 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would assume the base density to be = 1
We would convert 300 mg to g = 0.3 g
Applying the formula
Weight of base = (0.94 - (0.3 / 1.75)) * 1
Weight of base for 1 suppository = 0.7685
Weight of base for 25 suppositories = 0.7685 * 25 = 19.2 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would assume the base density to be = 1
We would convert 300 mg to g = 0.3 g
Applying the formula
Weight of base = (0.94 - (0.3 / 1.75)) * 1
Weight of base for 1 suppository = 0.7685
Weight of base for 25 suppositories = 0.7685 * 25 = 19.2 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would assume the base density to be = 1
We would convert 150 mg to g = 0.15 g
Applying the formula
Weight of base = (0.95 - (0.15 / 1.75)) * 1
Weight of base for 1 suppository = 0.8642
Weight of base for 10 suppositories = 0.8642 * 10 = 8.64 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would convert 150 mg to g = 0.15 g
Applying the formula
Weight of base = (0.95 - (0.15 / 1.75)) * 1.2
Weight of base for 1 suppository = 1.0370
Weight of base for 10 suppositories = 1.03704 * 10 = 10.37 g

Firstly lets solve with the displacement value method
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
Applying the formula
Weight of base = (2 - (0.2 / 2.5)) * 1
Weight of base for 1 suppository = 1.92
Weight of base for 10 suppositories = 1.92 * 20 = 38.4 g

Now lets solve for the weight of base required using the aaccurate method
When the strength of the drug is expressed as percentage we do not use displacement value method, rather we use:
Total weight of suppositories = weight per suppository * total amount of suppository
Total weight of suppositories = 2 * 20 = 40 g
To find total weight of active drug we say
10 % w/w means; 10 g of active drug in 100 g of suppositories
X g of active drug will be contained in 40 g of suppositories
cross multiply terms
X = 4 g of active drug
Weight of base required = Total weight of suppositories - total weight of active = 40 - 4 = 36 g

Percentage error incured if displacement value method is used = (Weight of base (from displacement value) - Weight of base (from actual method) ) / Weight of base from actual * 100 = (38.4 - 36) / 36 * 100 = 6.67 %

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would convert 400 mg to g = 0.4 g
Applying the formula
Weight of base = (0.94 - (0.4 / 1.5)) * 1.2
Weight of base for 1 suppository = 0.808
Weight of base for 10 suppositories = 0.808 * 25 = 20.2 g
Total proportion = 4 + 1 = 5
PEG 300 proportion = 1 / 5 * 20.2 = 4.04 g
PEG 4000 proportion = 4 / 5 * 20.2 = 16.16

In calculating the weight of base required for percentage suppository formulation displacement value method is not used
Rather we use: 7.5 % benzoicaine means 7.5 g in 100 g
Total weight of suppository = Weight of 1 suppository * Total number of suppositories = 1 * 25 = 25 g
If 7.5 g of benzocaine is contained in 100 g of benzocaine weight
X g of benzocaine would be contained in 25 g of suppositories
cross multiply terms
X = 1.875 g
Weight of base required = Total weight of suppositories - Total weight of active = 25 - 1.875 = 23.175 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would convert 50 mg to g = 0.05 g
Applying the formula
Weight of base = (0.90 - (0.05 / 1.6)) * 1.2
Weight of base for 1 suppository = 1.0425
Weight of base for 10 suppositories = 1.0425 * 15 = 15.64 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density
We would convert 50 mg to g = 0.05 g
Applying the formula
Weight of base = (0.90 - (0.05 / 1.3)) * 1.2
Weight of base for 1 suppository = 1.0338
Weight of base for 10 suppositories = 1.0338 * 8 = 8.27 g

We can assume a base density of 1 if unknown
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity supplied
We would convert 300 mg to g = 0.3 g
Applying the formula
Weight of base = (1 - (0.3 / 1.5)) * 1 * 10
Weight of base for 10 suppositories = 8.0 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity supplied
We would convert 500 mg to g = 0.5 g
Applying the formula
Weight of base = (1 - (0.5 / 1.5)) * 1.2 * 10
Weight of base for 10 suppositories = 8.0 g

Recall displacement value method is not used for percentage suppository formulations
Total weight of suppository = weight per 1 * total number of suppository
Total weight of suppository = 10 * 1 = 10 g
10 % w/w means 10 g of benzocaine in 100 g of benzocaine weight
10 g in 100 g of weight
X g would be contained in 10 g weight
X = 1 g
Weight of base required = Total weight - weight of active = 10 - 1 = 9 g of base

First we would determine the amount of base required for 10 * 1 suppositories taking ONLY purified water and wool alcohol into consideration (because they are not expressed as percentages)
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity supplied
Applying the formula
Weight of base = (1 - (0.1 / 1) - (0.1 / 1)) * 1 * 10
Base density also assumed to be 1
Weight of base = 8 g

Next we find the weight of the base for mercurochrome
1 % w/w means 1 g in 100 g of suppositories
Weight of 10 suppositories = 10 * 1 = 10 g
If 1 g of mercurochrome is contained 100 g of suppositories
X g of mercurochrome would be contained in 10 g of suppositories
cross multiply terms
X = 0.1 g
Weight of base needed = Weight of base (for purified water and wool alcohol) - weight of mercurochrome = 8 - 0.1 = 7.9 g

First we would determine the amount of base required for 10 * 1 suppositories for hydrocortisone
Converting 30 mg to g = 0.03 g
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity supplied
Applying the formula
Weight of base = 1 - (0.03 / 1.6) * 1 * 12
Base density also assumed to be 1
Weight of base = 11.775 g

Next we find the weight of the base for 5 % benzocaine
5 % w/w means 5 g in 100 g of suppositories
Weight of 10 suppositories = 12 * 1 = 12 g
If 5 g of benzocaine is contained 100 g of suppositories
X g of benzocaine would be contained in 12 g of suppositories
cross multiply terms
X = 0.604 g
Weight of base needed = Weight of base (for hydrocortisone) - weight of benzocaine = 11.75 - 0.604 = 11.17 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity dispensed
We would convert 250 mg to g = 0.25 g
Applying the formula
Weight of base = (0.95 - (0.25 / 1.5)) * 1.2 * 10
Weight of base for 10 suppositories = 9.4 g

Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity dispensed
We would convert 100 mg to g = 0.1 g
Applying the formula
Weight of base = (1 - (0.1 / 1.5)) * 1.2 * 10
Weight of base for 10 suppositories = 11.2 g

Firstly lets solve with the displacement value method
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity dispensed
Applying the formula
Weight of base = (1 - (0.1 / 5)) * 1 * 10
Weight of base for 10 suppository = 9.8 g

Now lets solve for the weight of base required using the aaccurate method
When the strength of the drug is expressed as percentage we do not use displacement value method, rather we use:
Total weight of suppositories = weight per suppository * total amount of suppository
Total weight of suppositories = 1 * 10 = 10 g
To find total weight of active drug we say
10 % w/w means; 10 g of active drug in 100 g of suppositories
X g of active drug will be contained in 10 g of suppositories
cross multiply terms
X = 1 g of active drug
Weight of base required = Total weight of suppositories - total weight of active = 10 - 1 = 9 g

Percentage error incured if displacement value method is used = (Weight of base (from displacement value) - Weight of base (from actual method) ) / Weight of base from actual * 100 = (9.8 - 9) / 9 * 100 = 8.9 %

Firstly lets solve with the displacement value method
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity dispensed
Applying the formula
Weight of base = (1 - (0.1 / 4)) * 1 * 10
Weight of base for 10 suppository = 9.75 g

Now lets solve for the weight of base required using the aaccurate method
When the strength of the drug is expressed as percentage we do not use displacement value method, rather we use:
Total weight of suppositories = weight per suppository * total amount of suppository
Total weight of suppositories = 1 * 10 = 10 g
To find total weight of active drug we say
10 % w/w means; 10 g of active drug in 100 g of suppositories
X g of active drug will be contained in 10 g of suppositories
cross multiply terms
X = 1 g of active drug
Weight of base required = Total weight of suppositories - total weight of active = 10 - 1 = 9 g

Percentage error incured if displacement value method is used = (Weight of base (from displacement value) - Weight of base (from actual method) ) / Weight of base from actual * 100 = (9.75 - 9) / 9 * 100 = 8.3 %

Firstly lets solve with the displacement value method
Weight of base = (Mould calibration - (drug dose (in g) / displacement value)) * base density * quantity dispensed
Applying the formula
Weight of base = (1 - (0.1 / 1)) * 1 * 10
Weight of base for 10 suppository = 9.8 g

Now lets solve for the weight of base required using the aaccurate method
When the strength of the drug is expressed as percentage we do not use displacement value method, rather we use:
Total weight of suppositories = weight per suppository * total amount of suppository
Total weight of suppositories = 1 * 10 = 10 g
To find total weight of active drug we say
10 % w/w means; 10 g of active drug in 100 g of suppositories
X g of active drug will be contained in 10 g of suppositories
cross multiply terms
X = 1 g of active drug
Weight of base required = Total weight of suppositories - total weight of active = 10 - 1 = 9 g

Percentage error incured if displacement value method is used = (Weight of base (from displacement value) - Weight of base (from actual method) ) / Weight of base from actual * 100 = (9 - 9) / 9 * 100 = 0 %

If the drug has a DV of 1 no error is incured using the displacement value method

We would prove an equation here and apply it for the rest of this series
Recall no of moles = mass (in g) / molecular weight

no of moles = milliequivalent (mEq) / valency

(1) Therefore mass (in g) / molecular weight = mEq / valency

Mass = ? (our unknown)
Molecular weight = 203.3 g/mol

Now lets solve for mEq
Recall a dilution was made 10 ml (V1) was taken
And diluted to 1000 ml (V2)
To give 0.3 mEq/ml (C2)
Applying the dilution formula we can find the milliequivalent concentration of Magnesium before the dilution

C1 * V1 = C2 * V2
C1 * 10 = 0.3 mEq/ml * 1000
C1 = 30 mEq/ml
So for every ml of solution we can find 30 mEq of Mg²⁺
Since 100 ml of the solution is required to be prepared we could say
30 mEq in 1 ml
X mEq would be in 100 ml
cross multiply terms
X = 3000 mEq

Lets get the valecncy
MgCl₂ Mg²⁺ + 2CL^-
The valency refers to the charge of Magnesium which in this case = 2
Note anhydrous water is not taken into consideration

Applying our formula (1)
Mass / mol.wt = mEq / valency
Mass = (3000 * 203.3) / 2 = 304,950 mg (value in mg because we worked with milliequivalent)
Mass = 305 g

Recall mass (in g) / molecular weight = mEq / valency see 9.1

Lets solve for mEq for all calcium salts available and sum up

Lets find the valecncy all calcium salts available
CaCl₂ Ca²⁺ + 2CL^-
The valency refers to the charge of Magnesium which in this case = 2
For all calcium salts the valency is always 2

Applying our formula (1)
For calcium chloride mEq = (260 * 2) / 147 = 3.54
For calcium gluconate mEq = (610 * 2) / 448.5 = 2.72
For calcium lactate mEq = (500 * 2) / 308.3 = 3.24
Total mEq for all calcium salts = 3.54 + 2.72 + 3.24 = 9.5

We would prove another equation here and apply it for the rest of this series
Recall no of moles = mass (in g) / molecular weight

no of moles = milliosmoles (mOsm) / no of osmotically active parts
(2) Therefore mass (in g) / mol.wt = mOsm / osmotically active parts

Lets solve for all mOsm in available salts and sum up

Lets find the number of osmotically active parts of disodium hydrogen phosphate
Na₂HPO₄.12H₂O 2Na⁺ + H₂PO₄^2-
There are 2 parts of sodium reacting with 1 part of hydrogen phosphate = 2 + 1 = 3

Lets find the number of osmotically active parts for potassium dihydrogen phosphate
KH₂PO₄ K⁺ + HPO₄^-
Number of osmotically active parts = 1 + 1 = 2

Applying our formula (2)
For disodium hydrogen phosphate = (2000 * 3) / 358.1 = 16.76
For potassium dihydrogen phosphate mOsm = (1000 * 2) / 136.1 = 14.69
Total mOsm in the solution = 16.76 + 14.69 = 31.45

Recall mass (in g) / mol.wt = mOsm / osmotically active parts

Lets solve for all mOsm in available salts and sum up

Lets find the number of osmotically active parts of sodium citrate
C₆H₅Na₃O₇ C₆H₅O₇ + 3Na⁺
There are 3 parts of sodium reacting with 1 part of citrate = 3 + 1 = 4

Lets find the number of osmotically active parts for sodium bicarbonate
NaHCO₃ Na⁺ + HCO₃^-
Number of osmotically active parts = 1 + 1 = 2

Applying our formula (2)
For sodium citrate = (500 * 4) / 294.1 = 6.8
For sodium bicarbonate = (1500 * 2) / 84.1 = 35.67
Total mOsm in the solution = 35.67 + 6.8 = 42.47

We would prove the final equation here and apply it for the rest of this series
Recall no of moles = milliequivalent (mEq) / valecncy

Recall also no of moles = milliosmoles (mOsm) / no of osmotically active parts

(3) Therefore milliequivalent (mEq) / valency = milliosmoles (mOsm) / osmotically active parts

Lets find the number of osmotically active parts and valency of magnesium chloride
MgCl₂ Mg²⁺ + 2Cl^-
There are 2 parts of chloride reacting with 1 part of magnesium = 2 + 1 = 3
The vlaency of magnesium and chloride is = 2

Lets find the number of osmotically active parts and valency for potassium chloride
KCl K⁺ + Cl^-
Number of osmotically active parts = 1 + 1 = 2
Valency of chloride and potassium = 1

Applying our formula (3)
For magnesium chloride = (12 * 3) / 2 = 18
For potassium chloride = (20 * 2) / 1 = 40
Total mOsm in the solution = 40 + 18 = 58

Notice how the molecular weight was ignored with this solution, this method is optimium when milliequivalents are given in place of mass in grams

Recall mass (in g) / molecular weight = mEq / valency

Mass = ? (our unknown)
Molecular weight = 147 g/mol

Now lets solve for mEq
Recall a dilution was made 10 ml (V1) was taken
And diluted to 1000 ml (V2)
To give 0.1 mEq/ml (C2)
Applying the dilution formula we can find the milliequivalent concentration of calcium before the dilution

C1 * V1 = C2 * V2
C1 * 10 = 0.1 mEq/ml * 1000
C1 = 10 mEq/ml
So for every ml of solution we can find 10 mEq of calcium
Given that 100 ml of the solution is required to be prepared we could say
10 mEq in 1 ml
X mEq would be in 100 ml
cross multiply terms
X = 1000 mEq

Lets get the valecncy
CaCl₂ Ca²⁺ + 2CL^-
The valency refers to the charge of calcium which in this case = 2
Note anhydrous water is not taken into consideration

Applying our formula (1)
Mass / mol.wt = mEq / valency
Mass = (1000 * 147) / 2
Mass = 73,500 mg (value in mg because we worked with mEq) or 73.5 g

Recall mass (in g) / molecular weight = mEq / valency see 9.1

Lets solve for mEq for all sodium salts available and sum up

Lets find the valecncy all sodium salts available
C₆H₅Na₃O₇ C₆H₅O₇^- + 3Na⁺
The valency refers to the charge of sodium which in this case = 3
NaHCO₃ Na⁺ + HCO₃^-
The valency here = 1

Applying our formula (1)
For sodium citrate = (500 * 3) / 294.1 = 5.1
For sodium bicarbornate = (1500 * 1) / 84.1 = 17.8

Total mEq for all sodium salts available = 5.1 + 17.8 = 22.9

Recall no of moles = mass (g) / mol.wt

0.1 mmol per 1 minute
X mmol per per 60 minutes (1 hour)
cross multiply terms
X = 60 * 0.1 = 6
6 mmol per 60 minutes
Y mmol per 360 minutes (6 hours)
Y = 36 mmol
Converting mmol to moles = 36 / 1000 = 0.036 moles

Applying the formula
0.036 = mass / 74.6
Mass = 2.68 g

Given that this is a dilution question we would apply the dilution formula
C₁V₁ = C₂V₂
C₁ = 2 mEq/ml
V₁ = 25 ml
C₂ = 5 * 10^-3M
V₂ = ? unkown

Given that the units for C₁ and C₂ are different we would calculate using two different techniques

Method 1 -- Converting 2 mEq/ml of C₁ to Molar concentration
Recall Molarity (M) = no of moles / volume
no of moles = mEq / valency
Therefore (4) Molarity (M) = mEq / valency * 1 / volume
Before applying equation (4) lets solve for mEq of SO₄^2-
2 mEq/ml means 2 mEq 1 ml
X ml would be contained in 25 ml
X = 50 mEq
Applying equation (4) Molarity = 50 / 2 * 1 / 25 (NB:The valency in this case is 2 because the charge on SO₄^2-) is 2
Molarity = 1 M
Applying our dilution formula : 1 * 25 = 5 * 10^-3 * V₂
V₂ = 5,000 ml 0r 5 L

Method 2 -- Converting 5 * 10^-3M of C₂to mEq/ml
We would apply equation (4) above also
5 * 10^-3 = mEq/ml / 2 * 1 / 1 (volume here = 1, because mEq is expressed per ml)
mEq/ml = 0.01
Applying the dilution formula: 2 mEq/ml * 25 = 0.01 mEq /ml * V₂
V₂ = 5000 ml Or 5 L

We simply applied the dilution formula but to do so we ensured both units are the same by converting one to the other (mEq/ml to Molar concentration and vice versa).
Both method checks out

Recall mass (in g) / mol.wt = mOsm / osmotically active parts

Lets solve for all mOsm in available salts and sum up

Lets find the number of osmotically active parts of sodium bicarbonate
NaHCO₃ Na⁺ + HCO₃^-
There is 1 part of sodium reacting with 1 part of bicarbonate = 1 + 1 = 2

Lets find the number of osmotically active parts for calcium chloride
CaCl₂ Ca²⁺ + 2Cl^-
Number of osmotically active parts = 1 + 2 = 3
For sodium chloride : NaCl Na⁺ + Cl^-
No of osmotically active part = 1 + 1 = 2
For magnesium chloride: MgCl2 Mg²⁺ + 2Cl^-
No of osmotically active parts = 1 + 2 = 3

Applying our formula (2)
For sodium bicarbonate = (150 * 2) / 84.1 = 3.57
For calcium chloride = (300 * 3) / 147 = 6.12
For sodium chloride = (180 * 2) / 58.5 = 6.15
For magnesium chloride = (60 * 3) / 203.3 = 0.89
Total mOsm in the solution = 3.57 + 6.12 + 6.15 + 0.89 = 16.73

From the Mol.wt we could say
98.07 g of sulphuric acid in 1000 ml is contained in 1 M solution
X g of sulphuric acid would be contained in 0.02 M solution
Cross multiply terms
X = 1.9614 g in 1000 ml

1000 ml of the preparation would require 1.9614 g
500 ml would require 0.9807 g

Dilute sulphuric acid contains 10 % w/w
10 g of sulphuric acid in 100 g of sulphuric acid weight
0.9807 g would be contained in Y g of sulpuric acid weight
cross multiply terms
Y = 9.81 g

Finally we find the volume corresponding to this mass
Recall Volume = mass / density
Volume = 9.81 / 1.045 = 9.4 ml

Recall 0.03 % w/v means;
0.03 g of histamine base in 100 ml of mixture
X g of histamine base in 10 ml of mixture
Cross multiply terms
X = 0.003 g of histamine base or 3 mg

2.76 mg of histamine acid phosphate is equivalent to 1 mg of histamine base
Y mg of histamine phosphate acid would be equivalent to 3 mg of histamine base
Y = 3 * 2.76 = 8.28 mg of Histamine phosphate acid

0.05 mmol of potassium over 1 minute
60 * 0.05 over 1 hour (60 minutes)
3 * 12 (36 mmol) over 12 hours
Converting to mol = 36 / 1000 = 0.036 moles

Mass = no of moles * mol.wt = 0.036 * 74.6 = 2.68 g

For sodium citrate
C₆H₅Na₃O₇ C₆H₅O₇^- + 3Na⁺
294.1 mg sodium citrate requires 3 mmoles of sodium
1000 mg of sodium citrate would require X mmoles of sodium
cross multiplying terms
X = 10.2 mmoles

For sodium carbornate
Na₂CO₃ 2Na⁺ + CO₃^2-
106 mg sodium carbonate requires 2 mmoles of sodium
750 mg will require Y mmoles of sodium
Y = 15.09 mmoles

Total mmoles for all sodium salts available = 15.09 + 10.2 = 25.3

Step 1 apply the dilution formula to derive M1, the concentration that would yeild 0.1 M
M1 * V1 = M2 * V2
M1 * 5 = 0.1 * 250
M1 = 5 M

Step 2;
We establish the molecular relationship between FeSO₄ and Fe²⁺
FeSO₄.7H₂O Fe²⁺ + SO₄^2-
278 mg of ferrous sulphate requires 1 M of Fe²⁺
Y mg of ferrous sulphate would require 5 M of Fe²⁺
cross multiply terms
Y = 1390 mg in 1000 ml
The mass here is in 1000 ml because molarity is usually expressed in 1000 ml
(NB: Molarity is defined as the number of moles of solute per litre of solution)

To find the mass in 100 ml it becomes
1390 mg in 1000 ml
Y mg in 100 ml
cross multiply terms
Y = 139 mg

Vamin N contains 20 mmol of potassium per litre (note mmol/L is equivalent to mEq/L)
Given that the preparation requires 25 mmol/L (or mEq/L) we need to prepare only an additional 5 mEq/L
98.14 mg of CH₃COOK requires 1 mEq
X mg of CH₃COOK will require 5 mEq
X = 490.6 mg of CH₃COOK
The injection contains 2450 mg in 5 ml
490.6 mg would be contained in Y ml
Cross multiply terms
Y = 1 ml

Step 1; Find the total dose based on the weight
12 mg per kg
69 kg = 12 * 69 = 828 mg

Step 2; Find the mEq corresponding to the dose
Recall mass (in g) / molecular weight = mEq / valency

Lets find the valecncy of sodium
C₁₅H₁₅C₁₂NaO₈ C₁₅H₁₅C₁₂NaO₈ + Na⁺
The valency refers to the charge of sodium which in this case = 1
Applying the formula
(828 * 1) / 445.2 = 1.859

Step 3; Find the total mEq for all 4 doses
1 dose = 1.859
4 doses (for every 6 hours in 1 day) = 4 * 1.859 = 7.4 mEq

Recall mass (in g) / molecular weight = mEq / valency see 9.1

Lets solve for mEq for all sodium salts available and sum up

Lets find the valecncy all sodium salts available
C₆H₅Na₃O₇ C₆H₅O₇^- + 3Na⁺
The valency refers to the charge of sodium which in this case = 3
NaHCO₃ Na⁺ + HCO₃^-
The valency here = 1

Applying our formula (1)
For sodium citrate = (1000 * 3) / 294.1 = 10.2
For sodium bicarbornate = (750 * 1) / 84.1 = 8.914

Total mEq for all sodium salts available = 10.2 + 8.914 = 19.11 mEq

Recall mass (in g) / molecular weight = mEq / valency

Mass = ? (our unknown)
Molecular weight = 147 g/mol

Now lets solve for mEq
Recall a dilution was made 5 ml (V1) was taken
And diluted to 250 ml (V2)
To give 0.1 mEq/ml (C2)
Applying the dilution formula we can find the milliequivalent concentration of calcium before the dilution

C1 * V1 = C2 * V2
C1 * 5 = 0.1 mEq/ml * 250
C1 = 5 mEq/ml
So for every ml of solution we can find 5 mEq of calcium
Given that 50 ml of the solution is required to be prepared we could say
5 mEq in 1 ml
X mEq would be in 50 ml
cross multiply terms
X = 250 mEq

Lets get the valecncy
CaCl₂ Ca²⁺ + 2CL^-
The valency refers to the charge of calcium which in this case = 2
Note anhydrous water is not taken into consideration

Applying our formula (1)
Mass / mol.wt = mEq / valency
Mass = (250 * 147) / 2
Mass = 18,375 mg (value in mg because we worked with mEq) or 18.4 g

Recall(3) milliequivalent (mEq) / valency = milliosmoles (mOsm) / osmotically active parts

Lets find the number of osmotically active parts and valency of magnesium chloride
MgCl₂ Mg²⁺ + 2Cl^-
There are 2 parts of chloride reacting with 1 part of magnesium = 2 + 1 = 3
The vlaency of magnesium and chloride is = 2

Lets find the number of osmotically active parts and valency for potassium chloride
KCl K⁺ + Cl^-
Number of osmotically active parts = 1 + 1 = 2
Valency of chloride and potassium = 1

Applying our formula (3)
For magnesium chloride = (24 * 3) / 2 = 36
For potassium chloride = (10 * 2) / 1 = 20
Total mOsm in the solution = 36 + 20 = 56

Notice how the molecular weight was ignored using this method, this method is optimium when milliequivalents are given in place of mass in grams

Recall mass (in g) / mol.wt = mOsm / osmotically active parts

Lets solve for all mOsm in available salts and sum up

Lets find the number of osmotically active parts of sodium citrate
C₆H₅Na₃O₇ C₆H₅O₇ + 3Na⁺
There are 3 parts of sodium reacting with 1 part of citrate = 3 + 1 = 4

Lets find the number of osmotically active parts for sodium bicarbonate
NaHCO₃ Na⁺ + HCO₃^-
Number of osmotically active parts = 1 + 1 = 2

Applying our formula (2)
For sodium citrate = (1000 * 4) / 294.1 = 13.6
For sodium bicarbonate = (750 * 2) / 84.1 = 17.83
Total mOsm in the solution = 17.83 + 13.6 = 31.4

Recall (2) mass (in g) / mol.wt = mOsm / osmotically active parts

Lets solve for all mOsm in available salts and sum up

Lets find the number of osmotically active parts of disodium hydrogen phosphate
Na₂HPO₄.12H₂O 2Na⁺ + H₂PO₄^2-
There are 2 parts of sodium reacting with 1 part of hydrogen phosphate = 2 + 1 = 3

Lets find the number of osmotically active parts for potassium dihydrogen phosphate
KH₂PO₄ K⁺ + HPO₄^-
Number of osmotically active parts = 1 + 1 = 2

Applying our formula (2)
For disodium hydrogen phosphate = (1195 * 3) / 358.1 = 10
For potassium dihydrogen phosphate mOsm = (454 * 2) / 136.1 = 6.67
Total mOsm in the solution = 6.67 + 10 = 16.7 mOsm

Recall mass (in g) / molecular weight = mEq / valency see 9.1

Lets solve for mEq for all calcium salts available and sum up

Lets find the valecncy all calcium salts available
CaCl₂ Ca²⁺ + 2CL^-
The valency refers to the charge of Magnesium which in this case = 2
For all calcium salts the valency is always 2

Applying our formula (1)
For calcium chloride mEq = (130 * 2) / 147 = 1.76
For calcium gluconate mEq = (1120 * 2) / 448.5 = 4.99
For calcium lactate mEq = (410 * 2) / 308.3 = 2.659
Total mEq for all calcium salts = 1.76 + 4.99 + 2.659 = 9.4 mEq

Recall mass (in g) / molecular weight = mEq / valency

Mass = ? (our unknown)
Molecular weight = 203.3 g/mol

Now lets solve for mEq
Recall a dilution was made 5 ml (V1) was taken
And diluted to 1000 ml (V2)
To give 0.1 mEq/ml (C2)
Applying the dilution formula we can find the milliequivalent concentration of magnesium before the dilution

C1 * V1 = C2 * V2
C1 * 5 = 0.15 mEq/ml * 1000
C1 = 30 mEq/ml
So for every ml of solution we can find 30 mEq of magnesium
Given that 100 ml of the solution is required to be prepared we could say
30 mEq in 1 ml
X mEq would be in 100 ml
cross multiply terms
X = 3000 mEq

Lets get the valecncy
MgCl₂ Mg²⁺ + 2CL^-
The valency refers to the charge on magnesium which in this case = 2
Note anhydrous water is not taken into consideration

Applying our formula (1)
Mass / mol.wt = mEq / valency
Mass = (3000 * 203.3) / 2
Mass = 304,950 mg (value in mg because we worked with mEq) or 305 g

Recall Molarity = no of moles / volume
For KH₂PO₄; 0.067 = no of moles / 50
No of moles = 3.35 moles

Recall no of moles = mOsm / osmotically active parts

Lets find the number of osmotically active parts of potassium hydrogen phosphate
KH₂HPO₄ K⁺ + H₂PO₄^2-
There is 1 parts of potassium reacting with 1 part of hydrogen phosphate = 1 + 1 = 2
3.35 = mOsm / 2
mOsm for KH₂PO₄ = 6.7

For Sodium phosphate;
Molarity = no of moles / volume
No of moles = 50 * 0.067 = 3.35

No of osmotically active parts for Na₂PO₄.12H₂O
Na₂PO₄.12H₂O 2Na⁺ + PO₄^2-
No of osmotically active parts = 3
3.35 = mOsm / 3
mOsm for Na₂PO₄.12H₂O = 10.05
Total mOsm = 10.05 + 6.7 = 16.75 mOsm

Note SCE stands for; Saturated calomel electrode
Using the SCE method

Percentage of adjusting substance =

0.9 - (Percentage of drug A * SCE of drug A) + (Percentage of drug B * SCE of drug B)... + (Percentage of drug N * SCE of drug N)

SCE Adjusting substance

Percentage of adjusting substance = = 1.06 % w/v

1.06 g of boric acid is required to render 100 ml of the eye drop isotonic
25 ml of the eye drop would require 0.265 g of boric acid (or 265 mg)

Step 1 Find the percentage w/v of potassium
5 g of potassium in 1000 ml
X g in 100 ml
X = 0.5 g in 100 ml (0.5 % w/v)
Apply the SCE method see 10.1
Percentage of adjusting substance = = 3.25 % w/v

3.25 g of glucose is required to render 100 ml of the solution isotonic
1000 ml (1 L) of the solution would require 32.5 g of glucose

Step 1 Find the percentage w/v of potassium
2.98 g of potassium in 1000 ml
X g in 100 ml
X = 0.298 g in 100 ml (0.298 % w/v)
Step 2 Apply the SCE method see 10.1
Percentage of adjusting substance = = 4.2095 % w/v

4.2095 g of glucose is required to render 100 ml of the solution isotonic
1000 ml (1 L) of the solution would require 42.1 g of glucose

Applying the SCE method see 10.1
Percentage of adjusting substance = = 0.81 % w/v

0.81 g of boric acid is required to render 100 ml of the eye drop isotonic
25 ml of the eye drop would require 0.2025 g of boric acid (or 202.5 mg)

Applying the C_iso method




0.65 g of sodium chloride is required to render 100 ml of the eye drop isotonic
15 ml of the eye drop would require 0.0975 g of sodium chloride (or 97.5 mg)

Applying the FD_1% method

Percentage of adjusting substance =

Percentage of adjusting substance = = 1.05 % w/v
1.05 g of adjusting substance is needed for 100 ml of eye drop
30 ml of the eye drop would require 0.315 g of glycerol (or 315 mg)

Applying the FD_1% method Percentage of adjusting substance = = 1.29 % w/v

1.29 g of boric acid is required to render 100 ml of the eye drop isotonic
25 ml of the eye drop would require 0.3225 g of boric acid (or 322.5 mg)

Applying the C_iso method
see 10.5

0.416 g of sodium chloride is required to render 100 ml of the eye drop isotonic
20 ml of the eye drop would require 0.0833 g of NaCL (or 83 mg)

Applying the SCE method see 10.1
Percentage of adjusting substance = = 0.622 % w/v

0.622 g of NaCL is required to render 100 ml of the eye drop isotonic
25 ml of the eye drop would require 0.1555 g of NaCL(or 155.5 mg)

Applying the SCE method see 10.1
Percentage of adjusting substance = = 0.538 % w/v

0.538 g of NaCL is required to render 100 ml of the eye drop isotonic
25 ml of the eye drop would require 0.1345 g of NaCL(or 134.5 mg)

Applying the C_iso method
] = 3.96 % w/v

3.96 g of glucose is required to render 100 ml of the eye drop isotonic
25 ml of solution would require 0.99 g (or 99 mg) of glucose

Applying the FD_1% method Percentage of adjusting substance = = 3.72 % w/v

3.72 g of anhydrous glucose is required to render 100 ml of the solution isotonic
1 L (1,000 ml) of the solution would require 37.2 g

Applying the FD_1% method Percentage of adjusting substance = = 0.479 % w/v

0.479 g of sodium chloride is required to render 100 ml of the solution isotonic
250 ml of the solution would require 1.198 g

Applying the SCE method see 10.1
Percentage of adjusting substance = = 0.25 % w/v

0.25 g of NaCL is required to render 100 ml of the solution isotonic

Applying the C_iso method
see 10.5

0.4869 g of sodium chloride is required to render 100 ml of the solution isotonic
50 ml of the solution would require 0.24345 g of NaCL (or 243 mg)

The Henderson-Hasselbalch equation
pH = pKa + log

[

Salt

Acid

]

Step 1 Find the number of moles of citric acid and sodium citrate
Recall No of moles = mass / molecular weight
For Citric acid 1 % = 1 g in 100 ml, therefore 14.2 ml would contain 0.142 g
No of moles of citric acid = 0.142 / 192.1 = 0.00074
For Sodium citrate 1.5 % = 1.5 g in 100 ml therefore 35.8 ml would contain 0.537 g
No of moles = 0.537 / 294.1 = 0.00182

Step 2 Apply the H-H equation
pH = 6.4 + log[0.00182 / 0.00074]
pH = 6.8

The Henderson-Hasselbalch equation
pH = pKa + log

[

dissociated

undissociated

]

The dissociated corresponds to the salt and the undissociated corresponds to the Acid
5.5 = 3.5 + log[dissociated / undissociated]
2 = log[dissociated / undissociated]
Antilog[2] = dissociated / undissociated
100 = dissociated / undissociated
dissociated = 100, undissociated = 1, or in ratio 100 : 1

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Step 1 Find the number of moles of acetic acid and sodium acetate
Recall No of moles = mass / molecular weight
For acetic acid 0.6 % = 0.6 g in 100 ml, therefore 10 ml would contain 0.06 g
No of moles of acetic acid = 0.06 / 60.05 = 0.000999
For Sodium acetate 1.36 % = 1.36 g in 100 ml therefore 50 ml would contain 0.68 g
No of moles = 0.68 / 136.1 = 0.0049963

Step 2 Apply the H-H equation
pH = 4.8 + log[0.0049963 / 0.000990]
pH = 4.8 + 0.699 = 5.5

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Step 1 Find the number of moles of citric acid and sodium citrate
Recall No of moles = mass / molecular weight
For citric acid 0.2 % = 0.2 g in 100 ml, given that novolume was given we would assume the final volume to be 100 ml
No of moles of citric acid = 0.2 / 210.1 = 0.000951
For Sodium citrate 1.25 % = 1.25 g in 100 ml
No of moles = 1.25 / 294.1 = 0.00425

Step 2 Apply the H-H equation
pH = 4.8 + log[0.00425 / 0.000951]
pH = 4.65 + 0.65 = 5.23 or 5.3

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Step 1 Find the number of moles of acetic acid and sodium acetate
Recall No of moles = mass / molecular weight
For citric acid 0.6 % = 0.6 g in 100 ml, 15 ml would contain 0.09 g of acetic acid
No of moles of acatic acid = 0.09 / 60.05 = 0.0014987
For Sodium acetate 1.36 % = 1.36 g in 100 ml, 60 ml would contain 0.816 g
No of moles = 0.816 / 136.1 = 0.00599

Step 2 Apply the H-H equation
pH = 4.8 + log[0.00599 / 0.0014987]
pH = 4.8 + 0.6 = 5.4

The Henderson-Hasselbalch equation
pH = pKa + log

[

dissociated

undissociated

]

The dissociated corresponds to the salt and the undissociated corresponds to the Acid
6 = 5.3 + log[dissociated / undissociated]
0.7 = log[dissociated / undissociated]
Antilog[0.7] = dissociated / undissociated
5 = dissociated / undissociated
dissociated = 5, undissociated = 1, or in ratio 100 : 1

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Applying the H-H equation
pH = 4.8 + log[0.2 / 0.1]
pH = 4.8 + 0.3 = 5.1

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Applying the H-H equation
pH = 9.24 + log[0.05 / 0.005]
pH = 9.24 + 1 = 10.24

The Henderson-Hasselbalch equation
pH = pKa + log

[

dissociated

undissociated

]

The dissociated corresponds to the salt and the undissociated corresponds to the Acid
5.76 = 4.76 + log[dissociated / undissociated]
1 = log[dissociated / undissociated]
Antilog[1] = dissociated / undissociated
10 = dissociated / undissociated
dissociated [salt] = 10, undissociated [acid] = 1, or in ratio 10 : 1

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Applying the H-H equation
pH = 4.2 + log[0.01 / 0.001]
pH = 4.2 + 1 = 5.2

The Henderson-Hasselbalch equation
pH = pKa + log

[

salt

acid

]

Note there is a decrease in the concentration of acetic acid due to the addition of 0.04 M NaOH, there is also an increase in the concentration of sodium acetate

Applying the H-H equation
Find first the new pH at increased/decreased concentration
pH = 4.76 + log[(0.2 + 0.04) / (0.2 - 0.04)]
pH = 4.76 + 0.176 = 4.93

Next find the old pH before the change in concentration
pH = 4.76 + log[0.2 / 0.2]
pH = 4.76 + 0 = 4.76

Find the Δ In pH
Change in pH = 4.93 - 4.76 = 0.176

t_90% = 0.105 / k

First we would find k given that we know t_1/2
t_1/2 = 0.693 / k
k = 0.693 / 5.5
k = 0.126 day^-1

t₉₀ = 0.105 / 0.126
t_90% = 0.8333 days
t_90%(in hours) = 0.8333 * 24 = 20 hours

t_90% = 0.105 / k

First we would find k given that we know t_1/2
t_1/2 = 0.693 / k
k = 0.693 / 44.8
k = 0.0154 hour^-1

t_90% = 0.105 / 0.0154
t_90% = 6.8 hours

t_90% = 0.105 / k

First we would find k given that we know t_1/2
t_1/2 = 0.693 / k
k = 0.693 / 22.4
k = 0.0309 hour^-1

t_90% = 0.105 / 0.0309
t_90% = 3.4 hours

Note! the maximum time of infusion t_10% corresponds to the shelf life t_90% t₉₀ = 0.105 / k

First we would find k given that we know t_1/2
t_1/2 = 0.693 / k
k = 0.693 / 3.33
k = 0.2081 day^-1

t_90% = 0.105 / 0.2081
t_90% = 0.505 days or 12.1 hours

t₉₀ = 0.105 / k

First we would find k given that we know t_1/2
t_1/2 = 0.693 / k
k = 0.693 / 72
k = 0.09625 day^-1

t_90% = 0.105 / 0.09625
t_90% = 1.1 days

log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time
log 4.2 * 10^-3= log 5 * 10^-3 -
K = 0.00872 month^-1

Next we solve for the final concentration after 70 % degredation
This concentration would be 70 % 0f the C_o
70/100 * 5 * 10^-3 = 3.5 * 10^-3

Next we find the time taken for this degredation
log 3.5 * 10^-3 = log 5 * 10^-3-
t = 40.09 months (value in months because k value was in months)

log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time
log 6.2 * 10^-3= log 8 * 10^-3 -
K = 0.01019 month^-1

Next we solve for the final concentration after 50 % degredation
This concentration would be 50 % 0f the C_o
50/100 * 8 * 10^-3 = 4 * 10^-3

Next we find the time taken for this degredation
log 4 * 10^-3 = log 8 * 10^-3-
t = 68 months (value in months because k value was in months)

Note that the time it takes for 50 % decomposition is the t_1/2
Therefore we could solve using;
t_1/2 = 0.693 / k
t_1/2 = 0.693 / 0.01019
t_1/2 = 68 months (this confirms our answer, answer in months because k value is the rate in months)

log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time
log 16.2 * 10^-3= log 25 * 10^-3 -
K = 0.005651 month^-1

Next we solve for the final concentration after 85 % degredation
This concentration would be 85 % 0f the C_o
85/100 * 25 * 10^-3 = 21.25 * 10^-3

Next we find the time taken for this degredation
log 21.25 * 10^-3 = log 25 * 10^-3-
t = 30 months (value in months because k value was in months)

t_1/2 = 0.693 / k
k = 0.693 / 6
K = 0.1155 week^-1
t₉₀ = 0.105 / 0.1155
t₉₀ = 0.909 weeks (or 0.909 * 7 = 6.4 days)

t_1/2 = 0.693 / k
k = 0.693 / 90
K = 0.0077 day^-1
t₉₀ = 0.105 / 0.0077
t₉₀ = 13.64 days or 13 days, 15 hours, 16 minutes

t_1/2 = 0.693 / k
k = 0.693 / 125
K = 0.005544 day^-1
t₉₀ = 0.105 / 0.00554
t₉₀ = 19 days

log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time

If 25 % of its potency is lost after 12.5 years we can safely assume it retains 75 % of its potency
75 % = C_f
100 % = C_o
log 75 = log 100 -
K = 0.0230 year^-1
t₉₀ = 0.105 / 0.0230
t₉₀ = 4.57 years

t₉₀ = 0.105 / k
7 = 0.105 / k
k = 0.015 days^-1
log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time

log C_f = log 250 -
log C = 2.007
C = 10^2.007
C = 101.62 mg/5 ml

If 15 % is lost after 106 weeks then we can safely assume, the product retains 85 %
log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time

log 85 = log 100 -
k = 0.001533 week^-1
t₉₀ = 0.105 / 0.001533
t₉₀ = 68.5 weeks

log C_f = log C_o - [ (k * t) / 2.303 ]

Where C_f = final concentration
C_o = Initial concentration
k = degradation rate
t = time

log 1.80 = log 2.50 -
k = 0.05447 week^-1
t₉₀ = 0.105 / 0.05447
t₉₀ = 1.92 months or 57 days, 14 hours, 24 minutes

Step 1; Setup the molecular relationship between molecular weight and Molarity to find the weight contained in 0.05 M
1 M solution of HCl === 36.5 g of HCL in 1000 ml
1 M solution of HCL === 73 g of HCL in 2000 ml
0.05 M solution of HCL === X g of HCL in 2000 ml
X = 3.65 g of HCl

Step 2; Find the mass of dilute HCL 3.65 g of HCL would be contained in
Recall dilute HCL contains 10 % w/w of HCL
10 g of HCL in 100 g of HCL solution
3.65 g of HCL would be contained in Y g of solution
Y === (3.65 * 100) / 10
Y === 36.5 g

Step 3; Find the volume the weight of dilute HCL would occupy
Density = Mass / Volume
Volume = Mass / Density = 36.5 / 1.04 = 34.9 ml

Step 1; Setup the molecular relationship between Chlorhexidine gluconate and Chlorhexidine
C₁₂H₃₀C₁₂N₁₀.2C₆H₁₂O₇ 2C₆H₁₂O₇ + C₁₂H₃₀C₁₂N₁₀
Chlorhexidine gluconate consists of 1 Chlorhexidine molecule and 2 gluconic acid molecules

Step 2; Find the molecular weight of chlorhexidine
Recall molecular weight of chlorhexidine gluconate = 897.8
Mol.weight of chlorhexidine = ?
Mol.weight of gluconic acid = 196.19
Mol.weight of chlorhexidine = mol.weight of chlorhexidine gluconate - (2 * mol.weight of gluconic acid)
The 2 factor multiplier for gluconic acid comes from the molecular relationship in which gluconic acid is 2
Mol.weight of chlorhexidine = 897.8 - 2(196.19)
Mol.weight of chlorhexidine = 505.42

Step 3; Find the weight of chlorhexidine in 20 g of chlorhexidine gluconate by considering the reactants molecular weight as ratios
Weight of chlorhexidine = mol.weight of chlorhexidine / mol.weight of chlorhexidine gluconate * 20
Weight of chlorhexidine = 505.42 / 897.8 * 20
Weight of chlorhexidine = 11.26 g

Setup the molecular relationship between Lithium carbonate and lithium element
Li₂CO₃ 2Li⁺ + CO₃^2-
74 g of Li₂CO₃ yeilds 14 g of lithium
250 mg (0.25 g) of Li₂CO₃ would yeild X g of lithium
X = 14 / 74 * 0.25
X = 0.0473 g or 47.3 mg

Setup the molecular relationship between sodium flouride and flourine element
NaF F^- + Na⁺
42 g of NaF would yeild 19 g of Flouride
X g of NaF would yeild 500 mcg of Flouride
X = 42 / 19 * 0.0005
X = 1.105 * 10^-3 g or 1.11 mg

Setup the molecular relationship between Sodium p-amino Benzoic acid, p-amino benzoic acid and sodium bicarbonate

NH₂C₆H₄OOH + NaHCO₃ NH₂C₆H₄OONa + HCO₃^2-
137 g of PABA would yeild 159 g of sodium-PABA
X g of PABA 100 g of sodium-PABA
X = (137 * 100) / 159
X = 86.2 g

Setup the molecular relationship between Sodium p-amino Benzoic acid, p-amino benzoic acid and sodium bicarbonate

NH₂C₆H₄OOH + NaHCO₃ NH₂C₆H₄OONa + HCO₃^2-
84 g of sodium bicarbonate produces 159 g of sodium-PABA
X g of sodium bicarbonate would be required to produce 100 g of sodium-PABA
X = (84 * 100) / 159
X = 52.8 g

Step 1 : Setup the molecular relationship between acetic acid, potassium acetate and potassium bicarbonate to find the mass of acetic acid needed

CH₃COOH + KHCO₃ CH₃COOK + H₂CO₃
60 g of acetic acid yeilds 98 g of potassium acetate
X g of acetic acid would yeild 200 g of potassium acetate
X = (200 * 60) / 98
X = 122.45 g

Step 2 : Find the total weight of acetic acid the pure acetic acid is contained in
Recall acetic acid is 36 % w/w
36 g of acetic acid in 100 g of acetic acid weight
122.45 g of acetic acid would be contained in Y g of acetic acid weight
Y = 340.18 g

Find the volume of acetic acid the acetic acid weight would occupy
Density = mass / volume
Volume = 340.13 / 1.045 = 325.5 ml

Setup the molecular relationship between acetic acid, potassium acetate and potassium bicarbonate to find the mass of acetic acid needed

CH₃COOH + KHCO₃ CH₃COOK + H₂CO₃
100 g of potassium bicarbonate yeilds 98 g of potassium acetate
X g of potassium bicarbonate would yeild 200 g of potassium acetate
X = (200 * 100) / 98
X = 204.1 g

Setup the molecular relationship between citric acid and citric acid monohydrate

C₆H₈O₇.H₂O C₆H₈O₇ + H₂O
210 g of citric acid monohydrate yeilds 192 g of citric acid (this checks out given that 1 molecule of water is 18 g, 192 g + 18 g = 210 g)
X g of citric acid monohydrate would yeild 27.4 g of citric acid
X = (27.4 * 210) / 192
X = 30 g

Setup the molecular relationship between ferrous sulphate and ferrous ion

FeSO₄.2H₂O Fe²⁺ + SO₄^2-
193 g of ferrous sulphate yeilds 56 g of elemental iron
0.2 g of ferrous sulphate would yeild X g of elemental iron
X = (56 * 0.2) / 193
X = 0.05803 g per tablet
3 tablets = 0.05803 * 3
0.1740 g or 174 mg

Setup the molecular relationship between potassium chloride and potassium ion

KCL K⁺ + Cl^-
74.6 g of potassium chloride yeilds 39.1 g of potassium
2.98 g of potassium chloride would yeild X g of potassium
X = (2.98 * 39.1) / 74.6
X = 1.5619 g or 1561.9 mg

Recall no of moles = mass (in g) / mol.wt

Recall no of mmoles = mass (in mg) / mol.wt
No of mmoles = 1561.9 mg / 39.1 (we consider the atomic weight of potassium not KCL)
No of mmoles = 39.9 mmoles per 10 ml of solution (recall the totoal volume of injection was given as 10 ml)
10 ml of the injection contains 39.9 mmol of potassium
1 ml of the injection would contain Y mmol of potassium
Y = (1 * 39.9) / 10
Y = 3.99 mmoles or approx 4 mmoles per ml

Dose to prepare 5 mg/kg

5 mg per 1 kg
X mg 20 kg
X = (20 * 5) / 1
X = 100 mg per Dose
4 doses = 4 * 100 = 400 mg
7 days = 7 * 400 = 2800 mg

2 moles of aminophylline is equivalent 1 mole of theophylline
(180.2 * 2) = 360.2 g of theophylline would be equivalent to 456.6 g of aminophylline
2800 mg of theophylline would be equivalent to Y g of aminophylline
X = (2800 * 456.6) / 360.4
Y = 3540.8 mg or 3.54 g

First we determine how much carindacillin is needed over 12 hours period.

0.5 mg 1 kg
X mg 81.5 kg
X = 81.5 * 0.5 = 40.75 mg per minute
40.75 * 60 = 2445 mg per hour
2445 * 12 = 29340 mg or 29.34 g

Next we establish the molecular relationship between carindacillin and carindacillin sodium
C₂₆H₂₅O₆SNa C₂₆H₂₅O₆S + Na⁺
516.6 g of carindacillin sodium would yeild 496.3 g of carindacillin
X mg of carindacillin sodium would yeild 29.34 mg of carindacillin
X = (29.34 * 516.3) / 496.3
X = 30.5 g

Step 1 : Work out the amount of insulin dispensed if the patient takes 0.5 ml of U-80 insulin
80 units is contained in 1 ml of U-80 insulin
0.5 ml of U-80 insulin would contain X units of insulin
X = 80 * 0.5 = 40 units

Step 2 : Work out the amount of U-100 insulin required to deliver 40 units of insulin
100 units is contained in 1 ml of U-100 insulin
X ml of U-100 insulin would contain 40 units of insulin
X = 40 / 100 = 0.4 ml

Note the formula : Child dose =

BSA (Child)

BSA (Adult)

* adult dose (in mg)

Child dose = * 450
Child dose = 0.3611 * 450 = 162.5 mg

Find the amount of heparin contained per ml of saline
15,000 units of heparin is contained in 250 ml of normal saline
X unit of heparin would be contained in 1 ml of normal saline
X = 15000 / 250 = 60 units per ml of saline

Find the volume (in ml) required to deliver 20 units
60 units of heparin is contained in 1 ml of normal saline
20 units of heparin would be contained in Y ml of normal saline
Y = 20 / 60 = 0.33 ml

Find the number of drops contained in 0.33 ml
60 drops is contained in 1 ml
Z drops would be contained in 0.33 ml
Z = 60 * 0.33 = 19.8 drops or approx 20 drops

The total dose required is (12.5 * 2 * 3 ) + (18.75 * 2 * 2) + (25 * 2 * 2) = 250 mg
Each capsule contains 25 mg of indomethacin
250 mg of indomethacin would be contained in 250 / 25 = 10 capsules

Step 1 : Work out the amount of insulin dispensed if the patient takes 0.6 ml of U-60 insulin
60 units is contained in 1 ml of U-60 insulin
0.6 ml of U-60 insulin would contain X units of insulin
X = 60 * 0.6 = 36 units

Step 2 : Work out the amount of U-100 insulin required to deliver 36 units of insulin
100 units is contained in 1 ml of U-100 insulin
Y ml of U-100 insulin would contain 36 units of insulin
Y = 30 / 100 = 0.36 ml

5 mcg for every 1 kg
X mcg for 70.5 kg
X = 5 * 70.5 = 352.5 mcg per minute or 0.3525 mg per minute
200 mg of dopamine HCL is contained in 500 ml of solution
0.3525 mg of dopamine HCL would be contained in Y ml of solution
0.88 ml / minute

Note the formula : Child dose =

BSA (Child)

BSA (Adult)

* adult dose (in mg)

Child dose = * 100
Child dose = 0.4167 * 100 = 41.67 mg |Approx 42 mg|

The total dose required is (1 * 15 * 2) + (0.5 * 15 * 2) + (0.25 * 15 * 2) + (0.125 * 15 * 2) + (0.125 * 15 * 1) + (15 * 0.125 * 1) = 30 + 15 + 7.5 + 3.75 + 1.875 + 1.875 = 60 mg
1 tablet contains 5 mg of prednisolone and another 25 mg of prednisolone
2 * 25 mg and 2 * 5 mg would be required to make up the 60 mg

The total amount of dopamine HCL in the 5 ml ampoule is 40 * 5 = 200 mg
200 mg of dopamine HCL is contained in 5 ml of solution
200 mg of dopamine HCL would be contained in 500 ml of solution
200 / 500 = 0.4 mg per ml

The patient weight in kg is 154.5 / 2.2046 = 70 kg
1 mcg for every 1 kg
X mcg for 70 kg
X = 1 * 70 = 70 mcg per minute
100 mcg of glycerlyl trinitrate is contained in 1 ml of solution
70 mcg of glycerlyl trinitrate would be contained in Y ml of solution
Y = 70 / 100 = 0.7 ml per minute

The patient weight in kg is 80 kg
Step 1 : Find the amount of drug required over 4 hours
20 mcg for every 1 kg
X mcg for 80 kg
X = 20 * 80 = 1600 mcg per minute or 1.6 mg per minute
200 mg of dopamine hydrochloride is contained in 5 ml of solution

4 hours = 4 * 60 = 240 minutes
Dose over 4 hours = 1.6 * 240 = 384 mg

Step 2 : Find the volume of solution required
200 mg of dopamine hydrochloride is contained in 5 ml of solution
384 mg of dopamine hydrochloride would be contained in Y ml of solution
Y = 384 * 5 / 200 = 9.6 ml

0.05 mg is contained in 1 ml of elixir
X mg would be contained in 3 ml of elixir
X = 0.05 * 3 = 0.15 mg |approx 150 mcg|

Step 1 : Find the total drops needed
1 drop both eyes 3 times daily = 1 * 2 * 3 = 6 drops

Step 2 : Find the volume 6 drops will contain
20 drops is contained in 1 ml
6 drops would be contained in Y ml
Y = 6 / 20 = 0.3 ml

Step 3 : Find the dose of pilocarpine contained in 0.3 ml
4 g of pilocarpine nitrate is contained in 100 ml of solution
Z g of pilocarpine nitrate would be contained in 0.3 ml of solution
Z = (4 * 0.3) / 100 = 0.012 g |approx 12 mg|

0.75 mcg for every 1 kg
X mcg for 176.4 lb
X = 0.75 * 176.4 / 2.2046 = 60 mcg per minute
200 mcg of glycerly trinitrate is contained in 1 ml of solution
60 mcg of glycerly trinitrate would be contained in Y ml of solution
Y = 60 / 200 = 0.3 ml per minute

Step 1 : Find the total dose the patient is taking
10 ml every 4 hours
10 ml every 4 hours would be 60 ml per day

Step 2 : Find the volume of concentrated chloroform water in 60 ml mixture
2.5 % v/v means; 2.5 ml of chloroform is contained in 100 ml of chloroform solution
X ml of chloroform would be contained in 60 ml of chloroform solution
X = 2.5 * 60 / 100 = 1.5 ml of chloroform

Step 3 : Find the volume of chloroform in 1.5 ml of concentrated chloroform water
10 ml of chloroform is contained in 100 ml of concentrated chloroform water
Y ml of chloroform water would be contained in 1.5 ml of concentrated chloroform water
Y = 10 * 1.5 / 100 = 0.15 ml of chloroform

Step 1 : Express the concentration of the drug in percentage weight per volume
1 mg of terbutaline is contained in 1 ml of solution
X mg of terbutaline would be contained in 100 ml of solution
X = 1 * 100 = 100 mg of terbutaline in 100 ml of solution or |0.1 g of terbutaline in 100 ml of solution| or |0.1 % weight per volume|

Step 2 : Find the concentration of the drug after dilution (C2)
C1 * V1 = C2 * V2
0.1 * 2 = C2 * 250
C2 = 0.1 * 2 / 250 = 0.0008 g per ml or |0.8 mg per ml|

Step 3 : Express the new concentration in mcg per ml
0.0008 g of terbutaline is contained in 100 ml of Solution
Y g of terbutaline would be contained in 1 ml of solution
Y = 0.0008 / 100 = 0.000008 g or |0.008 mg per ml| or |8 mcg per ml|

Step 4 : Find the volume of terbutaline required to deliver 25 mcg
8 mcg of terbutaline (per minute) is contained in 1 ml of solution
25 mcg of terbutaline (per minute) would be contained in Z ml of solution
Z = 25 / 8 = 3.125 ml per minute

Step 5 : Find the number of drops contained in 3.125 ml
15 drops is contained in 1 ml
T drops would be contained in 3.125 ml
T = 15 * 3.125 = 46.875 drops or |approx 47 drops| per minute

Infusion rate = volume / time
1000 ml / 12 hours = 83.33 ml per hour
83.33 ml per hour = 83.33 / 60 = 1.39 ml per minute

Step 1 : Find the volume of solution in each dose
24 drops is contained in 1 ml
10 drops would be contained in X ml
X = 10 / 24 = 0.42 ml

Step 2 : Find the amount of calciferol in 0.42 ml
120 mg of calciferol is contained in 100 ml of solution
Y mg of calciferol is contained in 0.42 ml of solution
Y = 120 * 0.42 / 100 = 0.504 mg or |approx 0.5 mg|

Step 1 : Find the volume of amoxicillin per minute
125 ml for 60 minutes
X ml 1 minute
X = 125 / 60 = 2.08 ml per minute

Step 2 : Find the amount of amoxicillin in 2.08 ml
200 mg of amoxicillin is contained in 500 ml of solution
Y mg of amoxicillin would be contained in 2.08 ml of solution
Y = 8.132 mg

Step 1 : Find the total dose of tablet indomethacin the patient is taking per day
25 mg three times daily = 25 * 3 = 75 mg

Step 2 : Find how much indomethacin is applied by solution
1 % w/w means 1 g of indomethacin is contained in 100 g of solution
1000 mg of indomethacin is contained in 100 g of solution
Given that density = mass (g) / volume
0.800 = 100 / X
X = 100 / 0.800 = 125 ml
Volume of solution = 125 ml

Step 3 :Find the amount of indomethacin in 20 ml of solution
1000 mg of indomethacin is contained in 125 ml of solution
Y mg of indomethacin would be contained in 20 ml of solution
Y = 1000 * 20 / 125 = 160 mg
Total indomethacin consumed by the patient = 160 + 75 = 235 mg

Step 1 : Find the dose of carindacillin patient is taking
0.5 mg for 1 kg
X mg for 75 kg
X = 0.5 * 75 = 37.5 mg per minute

Step 2 : Find the dose of carindacillin over 12 hours
37.5 mg of carindacillin for every minute
Y mg of carindacillin over 12 hours (60 * 12)
Y = 37.5 * 60 * 12 = 27,000 mg or |27 g|

Step 3 : Establish the molecular relationship between carindacillin and carindacillin sodium
516.6 g of carindacillin sodium produces 493 g of carindacillin
Z g of carindacillin sodium so would produce 27 g of carindacillin
Z = 27 * 516.6 / 493 = 28.2 g

Step 1 : Find the total dose required
7.5 ml bd = 15 ml per day
15 * 4 = 60 ml for 4 days
Step 2 : Find the dose contained in 60 ml
2 mg for 5 ml
Y mg 60 ml
Y = 2 * 60 / 5 = 24 mg

Step 1 : Find the total dose required
250 mg Qid = 250 * 4 = 1000 mg per day
1000 * 10 = 10,000 mg for 10 days
Step 2 : Find the volume of mixture required
125 mg for 5 ml
10,000 mg Y ml
Y = 400 ml

2.5 ml of peppermint oil is contained in 60 drops
0.5 ml of peppermint oil would be contained in X drops
X = 60 * 0.5 / 2.5 = 12 drops

Step 1 : Find the rate of infusion in ml per minute
1500 ml for 24 hours = 1500 / 24 = 62.5 ml per hour
62.5 ml per hour = 62.5 / 60 = 1.04 ml per minute

Step 2 : Find the rate of infusion in drops per minute
1 ml = 25 drops
1.04 ml = 1.04 * 25 = 26 drops per minute

Step 1 : Find the rate of infusion in ml per minute
20 mcg for every 1 hour (60 minutes)
X mcg 1 minute
X = 0.33 mcg per minute

Step 2 : Find the volume 0.33 mcg would occupy
150 mcg is contained in 250 ml of Infusion
0.33 mcg would be contained in Y ml of Infusion
Y = 0.555 ml per minute

Step 3 : Find the rate of infusion in drops per minute
1 ml = 20 drops
0.555 ml = 20 * 0.555 = 11.11 drops per minute

1 kg == 2.20 lb
X kg == 187 lb
X = 187 / 2.2 = 85 kg

For 85 kg at a dose of 0.3 mg per kg;
Dose = 85 * 0.3 = 25.5 mg

1 m² of body surface area requires 18 mg
1.7 m² will require X mg
X = 18 * 1.72 = 30.96 mg

Dose interval = 3 times daily
Dose duration = 5 days
Total dose = dose * dose interval * dose duration
Total dose = 30.96 * 3 * 5 = 464.4 mg

1 m² of body surface area requires 0.8 mg
0.8 m² of body surface area will require X mg
X = 0.8 * 0.8 = 0.64 mg or |640 mcg|

1 dose = 6 ml
4 doses = 6 * 4 = 24 ml per day
Total volume = 100 ml
Number doses = Total volume / Each dose
Number of doses = 100 / 6 = 16.66 dose |approx 16 doses|
Round down because 0.6 is not a complete dose