Concept of Solution
There are several ways by which concentration of solution is quantitatively described: molality, molarity, mole fraction, ppm, mass % etc. Liquid- liquid solutions can be classified as ideal and non-ideal solution on the basis of Raoult's law. Suppose the molecules of two components are represented by A and B respectively. In pure components, the intermolecular attractive interaction will be of types A-A and B-B, whereas in the binary solution, A-B type of interactions will also be present. If intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, ideal solution is formed.
The properties of solution which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are called colligative properties.
The vapour pressure of pure benzene at 25°C is 0.950 bar. When a 0.5 g non-volatile, non-electrolyte solute is added to 39.0 g of benzene (molar mass 78 g mol−1), then the vapour pressure of solution becomes 0.945 bar. What is the molar mass of the solute?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 190 g mol−1
Lowering of vapour pressure is one of the colligative properties described in the passage: it depends only on the number of solute particles. When a non-volatile solute is dissolved, the vapour pressure of the solvent falls. By Raoult’s law, the relative lowering of vapour pressure equals the mole fraction of the solute:
\( \displaystyle \frac{p_1^{0}-p_1}{p_1^{0}}=x_2=\frac{n_2}{n_1+n_2} \)
For a dilute solution n2 is very small compared with n1, so n1 + n2 ≈ n1. Writing n = w/M gives:
\( \displaystyle \frac{p_1^{0}-p_1}{p_1^{0}}=\frac{w_2\times M_1}{M_2\times w_1} \)
\( \displaystyle M_2=\frac{w_2\times M_1\times p_1^{0}}{w_1\times(p_1^{0}-p_1)} \)
Here w2 = 0.5 g, w1 = 39.0 g, M1 = 78 g mol−1, p10 = 0.950 bar and p1 = 0.945 bar.
Lowering of vapour pressure:
\( \displaystyle p_1^{0}-p_1=0.950-0.945=0.005\ \mathrm{bar} \)
Substituting:
\( \displaystyle M_2=\frac{0.5\times78\times0.950}{39.0\times0.005} \)
\( \displaystyle M_2=\frac{37.05}{0.195}=190\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the solute is 190 g mol−1.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Relative Lowering of Vapour Pressure – “Example 1.6 The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol−1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?”