Pharmaceutical Calculations Demystified: Exam-Ready Solutions for Pharmacists
pharmaceutical calculations questions and answers


Note:: First try your best to solve these pharmaceutical calculations. If you find them difficult you can consult the step-by-step solutions at the end of the MCQ questions.



1) 17.2g of NaCl is taken and diluted into 2000 milliliters of solution. Calculate the molarity of the resulting solution?
a- 0.147 M
b- 0.172 M
c- 0.2 M
d- 2 M
e- 17.2 M

Solution:

To determine the molarity of a solution, we need to know the amount of solute (in moles) and the volume of the solution (in liters).

First, let's convert the given mass of NaCl into moles. We can do this by dividing the mass by the molar mass of NaCl:

molar mass of NaCl = 22.99 g/mol (for Na) + 35.45 g/mol (for Cl) = 58.44 g/mol

moles of NaCl = 17.2 g / 58.44 g/mol = 0.294 moles

Next, we need to convert the given volume of the solution from milliliters to liters:

volume of solution = 2000 mL / 1000 mL/L = 2.0 L

Now, we can calculate the molarity of the solution by dividing the moles of NaCl by the volume of the solution:

molarity = moles of NaCl / volume of solution
= 0.294 moles / 2.0 L
= 0.147 M

Therefore, the molarity of the solution is 0.147 M or 0.15 M.

Thus the correct answer is a.





2) A solution is given to us which is 6.35x10⁻⁶ Molar in concentration. If the molecular weight of the compound is 342.3, calculate ppm for this compound?
a- 2.175 ppm
b- 3.423 ppm
c- 6.350 ppm
d- 1200 ppm
e- 1250 ppm

Solution:

To calculate the parts per million (ppm) concentration of a solute in a solution, we need to know the mass of the solute present in a known mass or volume of the solution.

In this case, we can use the following formula to calculate the ppm concentration of the solute:

ppm = (mass of solute / mass of solution) x 10

where the mass of the solute is equal to the number of moles of the solute multiplied by its molecular weight, and the mass of the solution can be calculated from the volume of the solution and its density.

First, we can calculate the mass of the solute (in grams) present in 1 liter of the solution:

moles of solute = molarity x volume (in liters)
= 6.3510⁻⁶ M x 1 L
= 6.3510⁻⁶ moles

mass of solute = moles of solute x molecular weight
= 6.3510⁻⁶ moles x 342.3 g/mol
= 2.17540510⁻³ g

Next, we can calculate the mass of 1 liter of the solution, assuming it has the density of water (1 g/mL):

mass of solution = volume of solution x density
= 1 L x 1000 g/L
= 1000 g

Finally, we can substitute these values into the formula for ppm to obtain the concentration of the solute in parts per million:

ppm = (mass of solute / mass of solution) x 10
= (2.175405*10⁻³ g / 1000 g) x 10
= 2.175405 ppm

Therefore, the concentration of the solute in the solution is 2.175405 ppm.

Thus the correct answer is a.




3) The recommended dose of medication is 65 milligrams per kilogram of body weight. f a person weighs 36.2 pounds, then their dose of medication would be?
a- 650 mg
b- 362 mg
c- 1066 mg
d- 1250 mg
e- 250 mg

Solution:

To calculate the amount of dose for a 36.2 lbs person with a recommended dose of 65mg/kg, we first need to convert the weight of the person from pounds to kilograms:

2.2 lbs = 1 kg
36.2 lbs = 16.4 kg

Next, we can calculate the dose of the medication for the person using the recommended dose of 65 mg/kg:

dose = weight of person x dose per kg
= 16.4 kg x 65 mg/kg
= 1066 mg

Therefore, the amount of dose for the 36.2 lbs person with a recommended dose of 65mg/kg is 1066 mg.

Thus the correct answer is c.




4) Calculate the percentage of the drug present in the mixture, if 15 gm of the drug is dissolved in 150 gm of petroleum jelly?
a- 5.04%
b- 9.09%
c- 15%
d- 19%
e- 25%

Solution:

To calculate the percentage of drug present in the mixture, we need to know the mass of the drug and the total mass of the mixture. In this case, we are given that 15 gm of drug is dissolved in 150 gm of petroleum jelly, so the total mass of the mixture is:

Total mass = mass of drug + mass of petroleum jelly
= 15 gm + 150 gm
= 165 gm

The mass of the drug represents the portion of the mixture that is the drug, so we can calculate the percentage of drug present as:

Percentage of drug = (mass of drug / total mass) x 100%
= (15 gm / 165 gm) x 100%
= 9.09%

Therefore, the percentage of drug present in the mixture is 9.09%.

Thus the correct answer is b.




5) A 5 ml solution of 0.5% contains how many milligrams of the substance?
a- 5 mg
b- 10 mg
c- 20 mg
d- 25 mg
e- 50 mg

Solution:

To calculate what a 5 ml solution of 0.5% contains, we need to know what substance is being dissolved in the solution.

If we assume that it is a mass/volume percentage solution, then 0.5% would mean that there are 0.5 grams of the substance in 100 ml of the solution.

To determine what a 5 ml solution of 0.5% contains, we can use a proportion:

0.5 grams / 100 ml = x grams / 5 ml

Solving for x, we get:

x = (0.5 grams / 100 ml) x (5 ml)
x = 0.025 grams

Therefore, a 5 ml solution of 0.5% contains 0.025 grams (or 25 milligrams) of the substance.

Thus the correct answer is d.






6) The given concentration of a drug is 5mg per 1 ml. Calculate the concentration of the drug in 1tsp in micrograms?
a- 15,000
b- 25,000
c- 50,000
d- 75,000
e- 10,000

Solution:

The amount of drug in 1 mL is 5 mg. We can convert this to micrograms by multiplying by 1000:

5 mg = 5000 μg

1 tsp is equivalent to 5 mL. Therefore, the amount of drug in 1 tsp can be calculated as:

5000 μg/mL x 5 mL = 25,000 μg

So the amount of drug in 1 tsp is 25,000 micrograms.

Thus the correct answer is b.






7) Calculate the volume of acetone if the mass of acetone is 6 g and the density is 1.75 g/mL?
a- 3.43 mL.
b- 6.75 mL
c- 12 mL
d- 16 mL
e- 20 mL

Solution:

To calculate the volume of acetone, we need to use the formula:

density = mass / volume

We are given that the mass of acetone is 6 g and the density is 1.75 g/mL. Let's rearrange the formula to solve for volume:

volume = mass / density

Substituting the values we have:

volume = 6 g / 1.75 g/mL

Simplifying, we get:

volume = 3.43 mL

Therefore, the volume of acetone is 3.43 mL.

Thus the correct answer is a.






8) What will be the amount of NaCl in grams if the concentration is 18% NaCl in 100ml?
a- 4.24 grams
b- 9.86 grams
c- 18 grams
d- 25.6 grams
e- 36.5 grams

Solution:

If 18% NaCl is present in 100 mL, we can calculate the amount of NaCl in grams using the formula:

mass = concentration x volume

Here, the concentration is 18% which can be written as 0.18 in decimal form.

So, mass of NaCl = 0.18 x 100 mL = 18 grams

Therefore, 18% NaCl in 100 mL is 18 grams.

Thus the correct answer is c.



9) A solution is made by dissolving 17.52 grams of NaCl in 2000ml. What will be the molarity of this solution?
a- 0.05 M
b- 0.15 M
c- 0.25 M
d- 0.5 M
e- 1.0 M

Solution:

To calculate the molarity of the NaCl solution, we need to first convert the mass of NaCl to moles using its molar mass. The molar mass of NaCl is approximately 58.44 g/mol.

moles of NaCl = mass of NaCl / molar mass of NaCl
moles of NaCl = 17.52 g / 58.44 g/mol
moles of NaCl = 0.2995 mol

Next, we need to calculate the volume of the solution in liters:

volume of solution = 2000 ml = 2 L

Finally, we can calculate the molarity (M) of the solution using the formula:

Molarity = moles of solute / volume of solution in liters

Molarity = 0.2995 mol / 2 L
Molarity = 0.1498 M

Therefore, the molarity of the NaCl solution is 0.1498 M.

Thus the correct answer is b.





10) A mixture composed of 3 components is formulated as;
Component A = 15 grams
Component B = 30 grams
Component C = 50 grams
Determine the amount of component B present in 200g of the mixture?
a- 30 grams
b- 63.16 grams
c- 70.2 grams
d- 45.2 grams
e- 110 grams

Solution:

To determine the amount of component B present in 200g of the mixture, we first need to calculate the total amount of the mixture.

Total amount of the mixture = Amount of A + Amount of B + Amount of C

Total amount of the mixture = 15 g + 30 g + 50 g
Total amount of the mixture = 95 g

Next, we can calculate the fraction of the mixture that is made up of component B by dividing the amount of B by the total amount of the mixture:

Fraction of component B = Amount of B / Total amount of the mixture

Fraction of component B = 30 g / 95 g

Fraction of component B = 0.3158

So the fraction or part of component B in 200 g of the mixture is:

Amount of component B in 200 g = 200 g x 0.3158

Amount of component B in 200 g = 63.16 g

Therefore, 63.16 grams of component B will be present in 200 g of the mixture.

Thus the correct answer is b.





11) What dose should be given to a 6-year-old child who weighs 44 lbs if the recommended dose of the drug is 0.5 mg/kg?
a- 0.002g
b- 0.006g
c- 0.010g
d- 0.050g
e- 0.100g

Solution:

We need to first convert the weight from pounds to kilograms:

2.2 lbs = 1kg approximately

44 lbs / 2.2 = 20 kg

So the child weighs 20 kg.

The dose of the drug is 0.5 mg/kg, so to find the dose for the child, we can multiply their weight by the dose:

Dose = weight x dose per kg
= 20 kg x 0.5 mg/kg
= 10 mg

Therefore, the dose of the drug that should be given to the 6-year-old child who weighs 44 lbs is 10 mg.

10mg = 0.010g

Thus the correct answer is c.





12) The dose of a drug is 0.6mg. How many doses are contained in 108 mg of the drug?
a- 16
b- 66
c- 160
d- 180
e- 216

Solution:

To calculate the number of doses contained in 96 mg of the drug, we need to divide the total amount of drug by the dose per unit:

Number of doses = Total amount of drug / Dose per unit

In this case, the dose per unit is given as 0.6 mg, and the total amount of drug is 96 mg:

Number of doses = 96 mg / 0.6 mg per dose

Number of doses = 160 doses

Therefore, there are 160 doses contained in 96 mg of the drug.

Thus the correct answer is d.




13) Converting 60 grams into grains gives us?
a- 9250 grains
b- 925 grains
c- 9.25 grains
d- 0.925 grains
e- 0.0925 grains

Solution:

To convert grams to grains, we can use the conversion factor

1 gram = 15.4324 grains.

Therefore, to convert 60 grams to grains:

60 grams x 15.4324 grains/gram = 925.944 grains

So 60 grams is equal to 925.944 grains.

Thus the correct answer is b.





14) Converting 2 pints 3 fluid ounces into ml will give us;
a- 1635 ml 
b- 1035 ml
c- 135 ml
d- 13.5 ml
e- 0.135 ml

Solution:

To convert pints and fluid ounces to milliliters, we need to know the conversion factors. 

One pint is equal to 473.176 milliliters, and 

One fluid ounce is equal to 29.5735 milliliters.

So, to convert 2 pints 3 fluid ounces into milliliters:

Convert 2 pints to milliliters:
2 pints x 473.176 mL/pint = 946.352 mL

Convert 3 fluid ounces to milliliters:
3 fluid ounces x 29.5735 mL/fluid ounce = 88.7205 mL

Add the two values to get the total:
946.352 mL + 88.7205 mL = 1035.0725 mL

Therefore, 2 pints 3 fluid ounces is equal to 1035.0725 milliliters 

Thus the correct answer is b.




15) Calculate the weight of 50 ml of Hydrochloric Acid if its density is 1.18g/ml?
a- 159 grams
b- 59 grams
c- 5.9 grams
d- 0.590 grams
e- 0.059 grams

Solution:

To calculate the weight of 50 ml of hydrochloric acid, we can use the density of the acid which is 1.18 g/ml.

The formula to calculate the weight of a liquid is:

Weight = Volume x Density

Substituting the given values, we get:

Weight = 50 ml x 1.18 g/ml

Weight = 59 g

Therefore, the weight of 50 ml of hydrochloric acid whose density is 1.18 g/ml is 59 grams.

Thus the correct answer is b.




16) A preparation contains 30 parts of coal tar and 15 parts of petrolatum in a mixture of 150 parts in total. What will be the concentration of petrolatum in 500ml of the mixture?
a- 15 parts
b- 30 parts
c- 50 parts
d- 150 parts
e- 300 parts

Solution:

To find the concentration of petrolatum in 500 ml of the mixture, we first need to calculate the proportion of petrolatum in the total mixture:

Proportion of petrolatum = parts of petrolatum / total parts of mixture

In this case, the mixture contains 30 parts of coal tar, 15 parts of petrolatum, and enough of an adequate substance to make 150 parts in total. So the total parts of mixture is 150.

Proportion of petrolatum = 15 / 150

Proportion of petrolatum = 0.1

So petrolatum makes up 0.1, or 10%, of the mixture.

To calculate the amount of petrolatum in 500 ml of the mixture, we can use the proportion of petrolatum we found earlier:

Amount of petrolatum in 500 ml = 0.1 x 500 ml

Amount of petrolatum in 500 ml = 50 ml

Thus the correct answer is c.





17) If a prescription order requires 30 g of concentrated Sulfuric acid (density is 1.8g/ml). What volume should the pharmacist measure?
a- 0.167 ml
b- 1.67 ml
c- 16.67 ml
d- 166.7 ml
e- 1667 ml

Solution:

To find the volume of concentrated sulfuric acid required for the prescription order, we can use the following formula:

Volume = Mass / Density

where:

Mass = 30 g (given in the prescription order)
Density = 1.8 g/ml (given in the problem)

Substituting the given values, we get:

Volume = 30 g / 1.8 g/ml

Volume = 16.67 ml

Therefore, the pharmacist should measure 16.67 ml of concentrated sulfuric acid for the prescription order.

Thus the correct answer is c.






18) What volume, in milliliters, of a 10% KCl solution (molecular weight 74.6) would contain 5.0 milliequivalents (mEq) of K+?
a- 0.373 ml
b- 3.730 ml
c- 0.670 ml
d- 6.70 ml 
e- 10.0 ml 

Solution:

To calculate the volume of a 10% KCl solution that contains 5.0 mEq of K+, we need to first calculate the number of moles of K+ in 5.0 mEq using the following formula:

mEq = moles x valence

where:

mEq = 5.0 mEq (given in the problem)
valence = 1 (since K+ has a valence of 1)

So, we have:

5.0 mEq = moles x 1

moles = 5.0 mEq / 1

moles = 0.005 moles of K+

Next, we need to calculate the mass of K+ in 0.005 moles of K+ using the molecular weight of KCl:

mass = moles x molecular weight

where:

moles = 0.005 moles (calculated above)
molecular weight = 74.6 g/mol (given in the problem)

So, we have:

mass = 0.005 moles x 74.6 g/mol

mass = 0.373 g of K+

Now we can use the concentration of the 10% KCl solution (i.e., 10 g/100 ml) to calculate the volume of solution needed to provide 0.373 g of K+:

Concentration = Mass of solute / Volume of solution

Rearranging this equation to solve for Volume, we get:

Volume = Mass of solute / Concentration

Substituting the values, we get:

Volume = 0.373 g / (10 g/100 ml)

Volume = 3.73 ml

Therefore, 3.73 ml of the 10% KCl solution would contain 5.0 mEq of K+.

Thus the correct answer is b.






19) What is the amount of KCl (with a molecular weight of 74.6) in grams that would be required for a prescription calling for 25 milliequivalents of potassium chloride?
a- 65 mg
b- 365 mg
c- 865 mg
d- 1.865 g
e- 2.860 g

Solution:

To calculate the mass of KCl required for a prescription calling for 25 mEq of potassium chloride, we can use the following formula:

Mass = mEq * MW / 1000

where:

mEq = 25 mEq (given in the prescription)
MW = 74.6 g/mol (molecular weight of KCl)
1000 = conversion factor from millimoles to moles

Substituting the given values, we get:

Mass = 25 mEq * 74.6 g/mol / 1000

Mass = 1.865 g

Therefore, 1.865 grams of KCl are required for the prescription calling for 25 mEq of potassium chloride.

Thus the correct answer is d.





20) If a radioisotope 32P has a half-life of 14.3 days and follows first-order kinetics, what is the decay constant (K)?
a- 0.048/day
b- 0.07/day
c- 0.097/day
d- 0.10/day
e- 0.184/day

Solution:

For a radioisotope that follows first-order kinetics, the decay constant (K) can be calculated using the following formula:

K = ln(2) / t1/2

where:

ln = natural logarithm
2 = the number representing the factor by which the radioisotope decays in one half-life
t1/2 = half-life of the radioisotope

Substituting the values given in the problem, we get:

K = ln(2) / 14.3 days

Using a calculator, we can evaluate this expression to get:

K ≈ 0.0484 per day

Therefore, the decay constant (K) of the radioisotope 32P is approximately 0.0484 per day.

Thus the correct answer is a.






21) How many grams of Glacial Acetic Acid (99.9% W/V) must be added to 1 gal Purified Water to prepare an irrigation solution containing 0.25% W/V Acetic Acid?
a- 1.24 g
b- 5.56 g
c- 9.47 g
d- 12 g
e- 25 g

Solution:

To prepare the 0.25% W/V acetic acid irrigation solution, we will first need to convert the volume of purified water from gallons to milliliters;

Convert gallons to milliliters:
1 gallon ≈ 3.785 L ≈ 3,785 mL

Next, we can use the formula for dilution:

C1 x V1 = C2 x V2

where:
C1 = concentration of glacial acetic acid = 99.9% w/v
V1 = volume of glacial acetic acid required
C2 = concentration of acetic acid in the irrigation solution = 0.25% w/v
V2 = final volume of the irrigation solution = 3,785 mL

Rearranging the formula and plugging in the values:

V1 = (C2 x V2) / C1

V1 = (0.25% x 3,785 mL) / 99.9%
V1 = 9.47 g

Approximately 9.47 grams of glacial acetic acid (99.9% W/V) must be added to 1 gallon of purified water to prepare an irrigation solution containing 0.25% W/V acetic acid.

Thus the correct answer is c.




22) A hospital clinic requests 2 pounds of 2% Hydrocortisone ointment. How many grams of 5% Hydrocortisone ointment could be diluted with White Petrolatum to prepare this order?
a- 16.2 g
b- 23.5 g
c- 123.4 g
d- 363 g
e- 445 g

Solution:

We are given that the hospital clinic requires 2 pounds of 2% Hydrocortisone ointment. Let's convert this to grams:

1 pound = 1 / 2.200 kg = 0.454 kg = 454 grams
2 pounds =  454 grams x 2 = 908 grams

Now, we can use the following formula to solve the problem:

C1 x V1 = C2 x V2

where:
C1 = concentration of the initial ointment = 5% w/w
V1 = volume of the initial ointment required = ? grams
C2 = concentration of the final ointment = 2% w/w
V2 = volume of the final ointment = 908 grams

Plugging in the values;
C1 x V1 = C2 x V2
5% x V1 = 2% x 908
V1 = (2% x 908) / 5%
V1 = 363.2 grams

Thus the correct answer is d.




23) What amount of glycerin in milliliters is required to formulate an ointment weighing 1 pound and containing 15% w/w glycerin? The density of glycerin used is 1.25 g/ml.
a- 6.2 ml
b- 18.2 ml
c- 22.4 ml
d- 24.2 ml
e- 28.6 ml

Solution:

To answer this question, we need to convert the weight percentage of glycerin to the actual weight of glycerin needed to prepare 1 lb of ointment.

1 pound = 1 / 2.200 kg = 0.454 kg = 454 grams

We know that the ointment contains 5% w/w glycerin, which means that for every 100 grams of ointment, there are 5 grams of glycerin.

To find the weight of glycerin needed for 1 lb of ointment (454 grams), we can set up a simple proportion:

5 grams glycerin / 100 grams ointment = x grams glycerin / 454 grams ointment

Solving for x, we get:

x = (5 grams glycerin / 100 grams ointment) * 454 grams ointment
x = 22.7 grams glycerin

Now that we know the weight of glycerin needed, we can use the density of glycerin to convert grams to milliliters:

density = mass / volume 

volume = mass / density

volume = 22.7 grams glycerin  / 1.25 g/ml 

volume = = 18.16 ml glycerin

Therefore, approximately 18.16 ml of glycerin would be needed to prepare 1 lb of ointment containing 5% w/w glycerin.

Thus the correct answer is b.





24) A pharmacist is requested to repackage 12 lbs of ointment into jars to be labeled 2 oz. each. How many jars can be filled with the total amount of ointment?
a- 6
b- 24
c- 48
d- 80
e- 96

Solution:

To determine the number of jars that can be filled with 12 lb of ointment to be labeled as 2 oz, we first need to convert pounds to ounces.

1 lb = 16 oz

So, 12 lb of ointment is equal to:

12 lb x 16 oz/lb = 192 oz

Next, we can divide the total amount of ointment by the amount of ointment per jar to find the number of jars:

192 oz ÷ 2 oz/jar = 96 jars

Therefore, 96 jars can be filled with 12 lb of ointment to be labeled as 2 oz. each.

Thus the correct answer is e.





25) What is the appropriate maintenance dose, based on a normal maintenance dose of 2mg/lb of body weight, for a 120 lb patient with an estimated creatinine clearance rate of 40 ml/min?
a- 26 mg
b- 76 mg
c- 96 mg
d- 160mg
e- 240mg

Solution:

To calculate the appropriate maintenance dose for this patient, you need to consider their creatinine clearance rate and adjust the dose accordingly.

First, let's convert the patient's weight from pounds to kilograms:

120 lbs * (1 kg / 2.2046 lbs) ≈ 54.4 kg

Now, let's assume that the normal creatinine clearance rate is 100 mL/min for simplicity. In this case, the patient's clearance rate is 40 mL/min, which is 40% of the normal rate (40 mL/min / 100 mL/min).

Next, calculate the normal maintenance dose for this patient:

Normal maintenance dose = 2 mg/lb * 120 lbs = 240 mg

Now, adjust the maintenance dose according to the patient's creatinine clearance rate:

Adjusted maintenance dose = Normal maintenance dose * (Patient's clearance rate / Normal clearance rate)
Adjusted maintenance dose = 240 mg * (40 mL/min / 100 mL/min)
Adjusted maintenance dose = 240 mg * 0.4
Adjusted maintenance dose = 96 mg

So, the maintenance dose for this 120lb patient with a creatinine clearance rate of 40 mL/min should be approximately 96 mg. 

Thus the correct answer is c.