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# 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.

2.17   The vapour pressure of water is $12.3 \; k Pa$ at $300 \; K.$  Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

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It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in 1000 g H2O.

Moles in 1000g of water = 55.55 mol.         (Since the molecular weight of H2O is 18)

Mole fraction of solute :

$\frac{1}{1+55.55} = 0.0177$

Applying the equation :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = x_2$

or                                       $\frac{12.3 - p}{12.3} = 0.0177$

or                                            $p = 12.083\ KPa$

Thus the vapour pressure of the solution is 12.083 KPa

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