Q.10% solution of non-volatile solute in water has a vapour pressure 740 mm at 373 K. Calculate the molar mass of the solute. OR What mass of ethylene glycol must be added to 5.5 kg of water to lower the freezing point from 0°C to −10°C? $K_f$ for water = 1.86 K kg mol$^{-1}$, molar mass of ethylene glycol = 62 g mol$^{-1}$.

Relative lowering of vapour pressure depends on the number of dissolved particles; Na3PO4 dissociates into 4 ions, the most among the options.

Lowering of vapour pressure is a colligative property, proportional to the total particle concentration (i x molarity). At 0.1 M:

• Urea: i = 1 → 0.1
• BaCl2: i = 3 → 0.3
• Na2SO4: i = 3 → 0.3 …

Both solutions contain the same number of moles of solute (0.2 mol) in the same amount of water, so their mole fractions of solute — and hence their relative lowering of vapour pressure — are equal. Correct option: (iv).

Concept. Relative lowering of vapour pressure (RLVP) is a colligative property: it depends on the number of solute particles, not their nature. By Raoult's law for a non-volatile solute,

$\frac{p^{0}-p_s}{p^{0}} = x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Why compare moles. Since both solutions use the same solvent quantity (1 L water), the one with more solute moles has the larger RLVP. So we just compare solute moles.

Steps.

• Urea: $n = \dfrac{12}{60} = 0.2\ \text{mol}$ (molar mass of urea, $\text{NH}_2\text{CONH}_2 = 60\ \text{g mol}^{-1}$).
• Sucrose: $n = \dfrac{68.4}{342} = 0.2\ \text{mol}$ (molar mass of sucrose, $\text{C}_{12}\text{H}_{22}\text{O}_{11} = 342\ \text{g mol}^{-1}$). …

Vapour pressure is not a colligative property; the colligative one is the RELATIVE LOWERING of vapour pressure.

A colligative property depends only on the number of solute particles present, not on their nature. The four colligative properties are:

• Relative lowering of vapour pressure
• Depression in freezing point
• Elevation in boiling point
• Osmotic pressure …

Adding a non-volatile solute to a solvent reduces the fraction of solvent molecules at the surface that can escape into vapour, so the vapour pressure of the solution is lower than that of the pure solvent.

According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solution (p) is p = p° × x_solvent, where p° is the vapour pressure of pure solvent and x_solvent < 1 is the mole fraction of solvent (since some of the moles are now solute). Because x_solvent is less than 1, p < p°, i.e. the vapour pressure is lowered on dissolving the solute.

…

Colligative properties depend only on the NUMBER of solute particles, not their identity: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, and osmotic pressure. Boiling point never falls when a non-volatile solute is added, so 'depression in boiling point' is not a real colligative property.

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lowered (Raoult's law). A lower vapour pressure means the solution needs to be heated to a HIGHER temperature to reach atmospheric pressure and boil, so the boiling point RISES (elevation), never falls.

…

The relative lowering of vapour pressure equals the mole fraction of the solute in the solution. Given the lowering is 0.0225, the mole fraction of the non-volatile solute is directly 0.0225, which corresponds to option (D).

Raoult's law states directly that the relative lowering of vapour pressure equals the solute's mole fraction.

According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure is given by:

$\frac{p°_{solvent} - p_{solution}}{p°_{solvent}} = x_{solute}$

…

Raoult's law directly equates the relative lowering of vapour pressure to the solute's mole fraction.

For a solution of a non-volatile solute in a volatile solvent, Raoult's law states:

(P0 - Ps)/P0 = x(solute) …

A non-volatile solute reduces the fraction of solvent molecules at the surface, lowering the rate of escape into vapour and hence the vapour pressure.

Vapour pressure arises from solvent molecules escaping from the liquid surface into the vapour phase. When a non-volatile solute is dissolved, some solute particles occupy positions at (or near) the surface, effectively reducing the fraction/number of solvent molecules present per unit area of the surface. This lowers the rate at which solvent molecules can escape into the vapour phase, so a smaller vapour pressure is need …