A pharmacist wishes to make a 250 mL solution of a 15 % w/v of a drug in Absolute Alcohol. How many milliltres of Absolute Alcohol would be needed ?
Note: specific gravity of Absolute Alcohol = 0.798 specific gravity of 15 % w/v of the drug in Absolute Alcohol = 0.952

Question

BlackTom AI · Accepted Answer

15% w/v means 15 g of drug per 100 mL of final solution, so for 250 mL the drug mass = 15/100 × 250 = 37.5 g.

1) Relevant knowledge points and concepts
- "% w/v" denotes grams solute per 100 mL of final solution (not solute per solvent).
- Specific gravity (SG) is the density relative to water; with water = 1.00 g/mL, density (g/mL) = SG.
- Total mass of the final solution = density × volume. Mass of solvent = total mass − mass of solute. Volume of solvent = mass of solvent ÷ solvent density.

2) Step-by-step reasoning and calculations
- Mass of drug required: 15 g/100 mL × 250 mL = 37.5 g.
- Density of final solution = SGfinal = 0.952 g/mL. Thus mass of 250 mL final solution:
  mass_solution = 0.952 g/mL × 250 mL = 238.0 g.
- Mass of absolute alcohol (solvent) needed:
  mass_alcohol = mass_solution − mass_drug = 238.0 g − 37.5 g = 200.5 g.
- Specific gravity of absolute alcohol = 0.798 g/mL, so volume of absolute alcohol:
  volume_alcohol = mass_alcohol ÷ density_alcohol = 200.5 g ÷ 0.798 g/mL = 251.253... mL ≈ 251.3 mL.

3) Intermediate steps shown above (all arithmetic displayed).

4) Why the correct answer is correct
Because the calculation uses the definition of w/v to find solute mass, uses the given solution SG to find total solution mass, subtracts solute mass to get solvent mass, then converts solvent mass to volume using the solvent SG. This yields 251.3 mL of absolute alcohol required.

5) Why the other options are incorrect
- c. 200.5 mL equals the mass of alcohol in grams (200.5 g) but neglects conversion to mL using alcohol density.  
- b. 214.9 mL, d. 175.3 mL, e. 154.5 mL result from incorrect or inconsistent use of densities or mixing up volume/mass; they do not follow the proper mass → volume conversion shown above.

6) Conclusion
Therefore the required volume of Absolute Alcohol is approximately 251.3 mL, which matches option a.

A pharmacist wishes to make a 250 mL solution of a 15 % w/v of a drug in Absolute Alcohol. How many milliltres of Absolute Alcohol would be needed ? Note: specific gravity of Absolute Alcohol = 0.798 specific gravity of 15 % w/v of the drug in Absolute Alcohol = 0.952Single choice

Similar Questions

How much water would be needed to prepare 500 mL of a saturated Potassium Iodide (KI) solution containing 100 g KI per 100mL of solution? (density of saturated solution = 1.7 g/mL, density of water = 1.00 g/mL)

Express a 1:40 solution of Sodium Hydroxide in terms of % w/v.

Please express the concentration term "1 in 2500" of a solution as a percentage. Assume we are referring to a solid dissolved in a liquid.

Express a 1:500 solution of Atropine Sulphate in terms of mg per mL.

T/F: The observations above suggest that the liquid in cup A could have been a very highly concentrated salt and sugar solution.

2.7 Calculate solution concentration

You have 5 liters of water with a concentration of 15 mg/L of Calcium dissolved in it. You leave the water in the sun and 2.5 liters of water evaporate. What is the concentration now?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler