The vapour pressure of water is 12.3 kPa at 300 K.
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

$\therefore$ Number of moles present in $1000 \mathrm{~g}$ 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, $p_{1}^{0}=12.3 \mathrm{kPa}$

Applying the relation, $\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=x_{2}$

$\Rightarrow \frac{12.3-p_{1}}{12.3}=0.0177$

⇒ 12.3 − p1 = 0.2177

⇒ p1 = 12.0823

= 12.08 kPa (approximately)

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

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