A 0.351 molal (m) aqueous solution of methanol (molar mass = 32.04 g/mol) is prepared. What is the concentration of methanol in percent mass/volume (%m/v)? Assume the density of water is 1.00 g/mL and that the volume change upon dissolving is negligible.

• A. 3.20 %m/v
• B. 1.13 %m/v
• C. 0.35 %m/v
• D. 0.11 %m/v

Answer: (B)

Verified step by step guidance

1. Understand the definition of molality (m): it is the number of moles of solute per kilogram of solvent. Here, molality = 0.351 mol/kg of water.
2. Calculate the mass of methanol using the molality and molar mass: mass of methanol (g) = molality (mol/kg) \times molar mass (g/mol) \times mass of solvent (kg). Since molality is per 1 kg of solvent, assume 1 kg of water for simplicity.
3. Calculate the total volume of the solution. Since the volume change is negligible and the density of water is 1.00 g/mL, the volume of the solvent (water) is approximately 1000 mL.
4. Calculate the percent mass/volume (%m/v) concentration using the formula: %m/v = (mass of solute in grams / volume of solution in mL) \times 100%. Use the mass of methanol from step 2 and the volume from step 3.
5. Express the final answer as %m/v, which represents grams of methanol per 100 mL of solution.