#### Question

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.

#### Solution

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

`x_2= 1/(1+55.56)= 0.0177`

Vapour pressure of water, `p_1^0 =12.3 kPa`

Applying the relation, `(p_1^0-p_1)/p_1^0 = x_2`

`=> (12.3- p_1)/12.3 = 0.0177`

`=>12.3 - p_1 = 0.2177`

= 12.08 kPa (approximately)

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

Is there an error in this question or solution?

Solution 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. Concept: Vapour Pressure of Liquid Solutions - Vapour Pressure of Liquid- Liquid Solutions.