Question 17

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol^{- 1}

∴ Number of moles present in 1000 g of water = 1000/18

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

x₂ = 1 / (1+55.56) = 0.0177.

It is given that,

Vapour pressure of water, p₁⁰= 12.3 kPa

Applying the relation, (P₁⁰- P₁) / P₁⁰= X₂

⇒ (12.3 - p₁) / 12.3 =0.0177

⇒ 12.3 - P₁ = 0.2177

⇒ p₁ = 12.0823

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

3 Comment(s) on this Question

2019-05-31 13:54:09

, mind-blowing

Ridham Rana 2019-05-28 10:44:24

Thank you for help SaralStudy.

Bhadawar 2019-04-04 20:46:15

Thanx for help