Molar mass of a non-volatile solute ( in g/mol) in its 10% by mass aqueous solution showing 2% decrease in its vapour pressure is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 98
By Raoult’s law, the relative lowering of vapour pressure of a solution of a non-volatile solute equals the mole fraction of the solute:
\( \displaystyle \frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2} \)
A 10% by mass solution has 10 g solute in 100 g solution, i.e. in 90 g water:
\( \displaystyle n_1 = \frac{90}{18} = 5\ \text{mol} \)
A 2% decrease means the relative lowering is 0.02:
\( \displaystyle 0.02 = \frac{n_2}{5 + n_2} \;\Rightarrow\; 0.98\,n_2 = 0.1 \;\Rightarrow\; n_2 = 0.102\ \text{mol} \)
\( \displaystyle M_2 = \frac{10}{0.102} = 98\ \text{g mol}^{-1} \)
Shortcut: the same law can also be written without approximation as \( \frac{p_1^0 - p_1}{p_1} = \frac{n_2}{n_1} \): \( \frac{0.02}{0.98} = \frac{10/M_2}{5} \), so M2 = 98 in one line. (The dilute-solution approximation n1 + n2 ≈ n1, 0.02 = n2/5, would give 100; with a 2% lowering n2 is not negligible.)