JEE Advanced 2022 Paper 2 · Q02 · Solutions & Colligative Properties

Question

An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35 °C. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. Vapour pressure of water at 35 °C is 60.000 mm of Hg. The number of ions present per formula unit of the ionic salt is _____.

SolveKar AI · Accepted Answer

Use Raoult's law with the van't Hoff factor.

Moles of water $= 1800/18 = 100$ mol. Mole fraction of solute particles (effective) $= \frac{i \cdot 0.1}{100 + i \cdot 0.1} \approx \frac{0.1\,i}{100}$ for small solute.

$\frac{P^\circ - P_s}{P^\circ} = \frac{60.000 - 59.724}{60.000} = \frac{0.276}{60} = 4.6 	imes 10^{-3}$.

So $\frac{0.1\,i}{100} = 4.6 	imes 10^{-3} \Rightarrow i = 4.6$.

For a salt dissociating into $n$ ions with degree of dissociation $\alpha = 0.9$:
$i = 1 + (n - 1)\alpha$
$4.6 = 1 + 0.9(n - 1)$
$3.6 = 0.9(n - 1) \Rightarrow n - 1 = 4 \Rightarrow n = 5$.

The salt yields 5 ions per formula unit.