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 = 0.952单项选择题

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