The vapour pressure of a solution is \(2985 Nm^{-2}\) when 5 g of non-electrolyte added into 100 g of water at particular temperature.
The vapour pressure of pure water at this temperature is \(3000 Nm^{-2}\). The molecular mass of the solute is.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 179
For a non-volatile, non-electrolyte solute, the relative lowering of vapour pressure equals the mole fraction of the solute (Raoult’s law):
\( \displaystyle \frac{p^\circ - p}{p^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \)
\( \displaystyle \frac{p^\circ - p}{p^\circ} = \frac{3000 - 2985}{3000} = \frac{15}{3000} = 0.005 \)
\( \displaystyle n_1 = \frac{100}{18} = 5.556\ \mathrm{mol},\qquad n_2 = \frac{5}{M} \)
\( \displaystyle n_2 = \frac{0.005\,n_1}{0.995} = \frac{0.005 \times 5.556}{0.995} = 0.02792\ \mathrm{mol} \)
\( \displaystyle M = \frac{5}{0.02792} = 179\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the solute is 179 g mol−1.
NCERT (2026-27 edition):
Relative Lowering of Vapour Pressure – “… Equation (1.24) can be written as
\( \displaystyle \frac{p_1^0 - p_1}{p_1^0} = x_2 \)
The expression on the left hand side of the equation as mentioned earlier is called relative lowering of vapour pressure and is equal to the mole fraction of the solute.”