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

Answer 1

The correct answer is: $12.08 kPa$
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$= \frac{1000}{18}$
$= 55.56\, mol$
Therefore, mole fraction of the solute in the solution is
$x_2 = \frac{1}{(1+55.56)}= 0.0177$
It is given that,
Vapour pressure of water,$p^0_1 = 12.3 kPa$
Applying the relation,$\frac{(p^o_1-p_1)}{p^o_1} = x_2$
$⇒\frac{(12.3-p_1)}{12.3}= 0.0177$
$⇒ 12.3 - p_1 = 0.2177$
$⇒ p_1 = 12.0823$
$= 12.08 kPa$ (approximately)
Hence, the vapour pressure of the solution is $12.08 kPa$