The mass fraction of urea (molar mass 60 g/mol) in an aqueous solution is 0.001. Calculate the relative lowering in vapour pressure of the solution.

0.01 %
0.02 %
0.03 %
0.04 %

Correct answer: 0.03 %

Solution

By Raoult's law, relative lowering of vapour pressure = x_solute. In 1000 g solution: mass of urea = 1 g, mass of water = 999 g. Moles of urea = 1/60 = 0.01667 mol. Moles of water = 999/18 = 55.50 mol. x_urea = 0.01667 / (0.01667 + 55.50) = 3.0 * 10⁻⁴. As a percentage: 3.0 * 10⁻⁴ * 100 = 0.030 % ≈ 0.03 %.

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