Solution of MCQ # 105 Pharmaceutical Calculation Problem

This is MCQ # 105 of 1000 Pharmacy MCQs

105. If a pharmacist incorporates 1 pint of Alcohol USP into 5 L of a mouthwash formula that was initially labeled as 10% V/V, what would be the percentage of alcohol present in the resulting mixture?

a- 14.34 %

b- 17.34 %

c- 19.68 %

d- 22.43 %

e- 24.40 %

Solution:

First of all, let's know something about Alcohol USP;

Alcohol USP (United States Pharmacopeia) refers to a specific grade of alcohol that is used for medicinal purposes, such as for use as an antiseptic or as a solvent in the manufacturing of pharmaceuticals. The concentration of alcohol in Alcohol USP is typically 95% ethanol by volume, which means that it contains 95% ethanol and 5% water.

So, Alcohol USP is 95% Alcohol.

In the question it is 1 pint of Alcohol USP, so we need to find out the amount of alcohol in 1 pint of Alcohol USP;

1 pint = 473.18 ml

So, (95ml / 100ml) x 473.18 ml = 449.521 ml

Thus, 1 pint of Alcohol USP contains 449.521 ml of alcohol.

Now we need to find out the amount of alcohol in 5 L of 10% V/V;

10% V/V means 10 ml alcohol is there in 100 ml of solution;

so, in 5 L (5000ml) of solution, how much alcohol?

(10 ml / 100 ml) x 5000 ml = 500 ml

so 500 ml of alcohol is there in 5 L of 10% V/V solution.

After adding 1 pint of Alcohol USP into 5 L of 10% V/V formula;

The total volume of alcohol becomes;

449.521 ml + 500 ml = 949.521 ml

The total volume of the mixture becomes;

473.18 ml + 5000 ml = 5473.18 ml

Now lets calculate the percentage of alcohol present in the resulting mixture;

( 949.521 ml / 5473.18 ml ) x 100 = 17.34%

Therefore, the correct answer is b.

Solution of MCQ # 830 Pharmaceutical Calculation Problem

This is MCQ # 830 of 1000 Pharmacy MCQs

830. How many grams of sodium chloride are there in 30 millimoles of NaCl, given the molecular weight of NaCl as 58.5?

a- 58.5

b- 30

c- 15

d- 2

e- 1.755

Solution:

To convert from millimoles to moles, we need to divide by 1000:

30 millimoles NaCl = 30/1000 moles NaCl = 0.03 moles NaCl

The mass of one mole of NaCl is 58.5 grams, so the mass of 0.03 moles of NaCl is:

0.03 moles NaCl × 58.5 grams/mole = 1.755 grams NaCl

Therefore, there are 1.755 grams of NaCl in 30 millimoles of NaCl.

Thus the correct answer is e.

Solution of MCQ # 826 Pharmaceutical Calculation Problem

This is MCQ # 826 of 1000 Pharmacy MCQs

826. 40 grams of a drug is added into 30 ml of a solvent. The density of the solvent is 1.5. What will be the concentration of the drug in the final product?

a- 26 %

b- 30 %

c- 40 %

d- 47 %

e- 60 %

Step-by-step solution:

To calculate the concentration of the drug in the final product, we first need to determine the mass of the solvent. We can use the formula:

mass = volume × density

Given the density of the solvent is 1.5 g/mL, we can calculate the mass of the solvent as follows:

mass = 30 mL × 1.5 g/mL = 45 g

Now, we have;

mass of the drug = 40 g

mass of the solvent = 45 g

To find the concentration of the drug, we need to find the mass fraction, which is the ratio of the mass of the drug to the total mass of the mixture. The formula for mass fraction is:

mass fraction = mass of drug / (mass of drug + mass of solvent)

Using our values:

mass fraction = 40 g / (40 g + 45 g) = 40 g / 85 g

So the mass fraction of the drug is:

mass fraction = 40 g / 85 g ≈ 0.4706

To express the concentration as a percentage, simply multiply the mass fraction by 100:

concentration (%) = mass fraction × 100 ≈ 0.4706 × 100 ≈ 47.06%

So the concentration of the drug in the final product is approximately 47.06%.

Thus the correct answer is d.

Solution of MCQ # 905 Pharmaceutical Calculation Problem

This is MCQ # 905 of 1000 Pharmacy MCQs

905. How much of a 2% stock solution of a certified green dye should be used to prepare 6000 milliliters of a mouthwash that is to contain 2:20,000 w/v of the certified green dye as the coloring agent?

a- 10 milliliters 

b- 20 milliliters 

c- 30 milliliters 

d- 60 milliliters 

e- 120 milliliters 

Simple Solution:

2:20,000 w/v means 0.01%

6000 milliliters / 2 x 0.01 = 30 milliliters 

Detailed Solution:

To calculate the amount of the 2% stock solution of certified green dye needed, we can use the dilutions formula:

C1V1 = C2V2

where,

C1 is the concentration of the stock solution, 

V1 is the volume of the stock solution to be used, C2 is the desired concentration of the dye in the mouthwash, 

and V2 is the final volume of the mouthwash.

First, we need to convert the desired concentration of the dye in the mouthwash from w/v (weight/volume) to percent. 

2:20,000 w/v means 2 grams of dye in 20,000 milliliters of the solution, or 0.01% (2/20,000 x 100%) concentration.

Now we can plug in the values:

C1 = 2%

V1 = ?

C2 = 0.01%

V2 = 6000 mL

2% V1 = 0.01% x 6000 mL

V1 = (0.01/2) x 6000

V1 = 30 mL

Therefore, we need to use 30 mL of the 2% stock solution of certified green dye to prepare 6000 milliliters of mouthwash that contain 2:20,000 w/v of the dye as the coloring agent.

Thus the correct answer for MCQ # 905 is c.

More Pharmaceutical Calculations Problems

Solution of MCQ # 917 Pharmaceutical Calculation Problem

This is MCQ # 917 of 1000 Pharmacy MCQs

917. At a pharmacy, we have a vial that contains 5ml of lidocaine 1% solution. How many milligrams of lidocaine are present in 2ml of the solution?

a- 0.01 mg

b- 0.02 mg

c- 2.5 mg

d- 10 mg
e- 20 mg 

Solution:

As we know;

lidocaine 1% solution means 1gm or 1000mg of lidocaine are there in 100ml of solution

As we have a 5ml vial available at our pharmacy, we have to calculate first the milligrams of lidocaine present in 5 ml.

1000mg in 100ml

?mg in 5ml

cross multiplying, we get

1000mg x 5ml = ?mg x 100ml

5000 / 100 = ?mg

So, 50mg of lidocaine are there in 5ml solution.

Now let's calculate milligrams of lidocaine in 2ml solution.

50mg in 5ml

?mg in 2ml

again cross multiplying;

?mg x 5ml = 50mg x 2ml

?mg = 100 / 5 = 20

So 20 milligrams of lidocaine are there in 2ml solution.

Thus the correct answer is e.

More Pharmaceutical Calculations Problems