A set of solutions is prepared using 180 g of water as the solvent and 10 g each of three different non-volatile non-electrolyte solutes A, B and C. What is the correct order of the relative lowering of vapour pressure of the solutions? (Molar masses: A = 100 g mol−1, B = 200 g mol−1, C = 10000 g mol−1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A > B > C
For a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute:
\( \displaystyle \frac{p_1^0-p_1}{p_1^0}=x_2=\frac{n_2}{n_1+n_2} \)
The solvent is the same in all three (180 g of water, n1 = 10 mol), so the solution with more moles of solute shows the larger relative lowering. With the same 10 g of each solute, the smaller the molar mass, the more moles:
A: \( \displaystyle n_2=\frac{10}{100}=0.1\,\mathrm{mol} \)
B: \( \displaystyle n_2=\frac{10}{200}=0.05\,\mathrm{mol} \)
C: \( \displaystyle n_2=\frac{10}{10000}=0.001\,\mathrm{mol} \)