JEE Advanced 2022 — Paper 2 — CHEM-I solution

JEE Advanced2022Paper 2CHEM-I

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. Vapor 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 _______.

Your answer

Solution

Let $x$ = number of ions per formula unit. For 90% dissociation, the van't Hoff factor is $i = 0.9 x + 0.1$ (90% dissociated giving $x$ ions each, plus 10% undissociated).

Moles of water $= 1800/18 = 100$ . Relative lowering of vapour pressure equals the mole fraction of solute (Raoult's law for dilute solutions, including the van't Hoff factor):

$\frac{60 - 59.724}{60} = \frac{i \cdot 0.1}{100 + 0.1}$

The left side equals $\frac{0.276}{60} = 0.0046$ . Hence $0.0046 \approx \frac{i \cdot 0.1}{100}$ , giving $i \cdot 0.1 = 0.46$ , so $i = 4.6$ .

From $i = 0.9 x + 0.1 = 4.6$ : $0.9 x = 4.5$ , so $x = 5$ .

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