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 mm Hg. The number of ions present per formula unit of the ionic salt is ______.

Answer 1

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 mm Hg. The number of ions present per formula unit of the ionic salt is 5.

Explanation:

Given: Moles of ionic salt = 0.1 mol

Mass of water = 1.8 kg = 1800 g

Vapour pressure of water P° = 60 mm Hg

Vapour pressure of solution P = 59.724 mm Hg

Degree of dissociation α = 90% = 0.9

Let the number of ions present per formula unit of ionic salt = x

van’t Hoff factor (i) = 0.9 × + 0.1 × l

As the relative lowering of V.P. = mole fraction of solute

∴ `(60 - 59.724)/60 = (i xx 0.1)/(1800/18 + 0.1)`

⇒ `0.0046 = (0.1 xx i)/(100 + 0.1)`

⇒ `0.0046 = ((0.9 x + 0.1) xx 0.1)/100`

⇒ `0.0046 = (0.09 x + 0.01)/100`

⇒ 0.46 = 0.09x + 0.01

⇒ 0.45 = 0.09x

⇒ x = 5