A 0.425 m aqueous solution of methanol (molar mass = 32.04 g/mol) is prepared. What is the concentration of methanol in %w/v (grams of solute per 100 mL of solution), assuming the density of the solution is approximately 1.00 g/mL?

• A. 0.425 %w/v
• B. 4.25 %w/v
• C. 1.36 %w/v
• D. 13.6 %w/v

Answer: (C)

Verified step by step guidance

1. Understand the meaning of molality (m): it is defined as moles of solute per kilogram of solvent. Here, the molality is given as 0.425 m, meaning 0.425 moles of methanol per 1 kg of water.
2. Calculate the mass of methanol in grams using the molality and molar mass. Use the formula: \(\text{mass of solute (g)} = \text{moles of solute} \times \text{molar mass (g/mol)}\). Since molality is moles per kg solvent, for 1 kg of water, moles of methanol = 0.425 moles.
3. Calculate the total mass of the solution by adding the mass of solvent (1000 g) and the mass of solute (from step 2).
4. Use the density of the solution (1.00 g/mL) to find the volume of the solution in milliliters: \(\text{volume (mL)} = \frac{\text{mass of solution (g)}}{\text{density (g/mL)}}\).
5. Finally, calculate the %w/v concentration, which is grams of solute per 100 mL of solution, using the formula: \(\%w/v = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100\).