If at a certain temperature the vapour pressure of pure water is 25 mm of Hg and that of a very dilute aqueous urea solution is 24.5 mm of Hg, the molality of the solution is:

1.11

The correct answer is option 3. 1.11 \(\text{mol kg}^{-1}\).

Molality: The number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent is known as molality. The definition of molarity, on the other hand, is based on a certain volume of solution.

\(Molality = \frac{n_{solute}}{M_{solvent}} × 1000\)

            \(= \frac{w_{solute} × 1000}{M_{solvent} × \text{Molar mass of solute}}\)

Urea is a non-volatile solute. Then according to Raoult's Law;

\(\frac{P_s}{P_o}= X_1\)

\(P_s = 24.5 \text{mm Hg}\); \(P_o = \text{ 25 mm Hg}\)

\(∴ X_1 = \frac{24.5}{25.0}  = 0.98\)

\( ∴ \text{Mole fraction of urea = }X_2 = 1 − X_1\)

                                            \(= 1 − 0.98 = 0.02\)

\(∴ Molality = \frac{\text{Number of moles of urea}}{\text{Mass of water in gram}} × 1000\)

                \(=\frac{0.02}{0.98 × 18} × 1000\)

                 \(≅ 1.11 \text{mol kg}^{-1}\)