# 2.19   A solution containing 30 g of non-volatile solute exactly in $\inline 90\; g$ of water has a vapour pressure of $\inline 2.8\; k\; Pa$ at $\inline 298 \; K.$. Further, $\inline 18 \; g$ of water is then added to the solution and the new vapour pressure becomes $\inline 2.9\; kPa$ at $\inline 298\; K.$. Calculate:             (i) molar mass of the solute

D Devendra Khairwa

In this question we will find molar mass of solute by using Raoult's law .

Let the molar mass of solute is M.

Initially we have 30 g solute and 90 g water.

Moles of water :

$\frac{90}{18} = 5\ mol$

By Raoult's law we have :-

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_2}{n_1+n_2}$

or                                        $\frac{p_w^{\circ} - 2.8}{p_w^{\circ}} = \frac{\frac{30}{M}}{5+\frac{30}{M}} = \frac{30}{5M+30}$

or                                         $\frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}$                                                                                ------------------------------   (i)

Now we have added 18 g of water more, so the equation becomes:

Moles of H2O  :

$\frac{90+18}{18} = 6\ mol$

Putting this in above equation we obtain :-

$\frac{p_w^{\circ} - 2.9}{p_w^{\circ}} = \frac{\frac{30}{M}}{6+\frac{30}{M}} = \frac{30}{6M+30}$

or                                  $\frac{p_w^{\circ} }{2.9} = \frac{6M +30}{6M}$                                                                                   -----------------------------------(ii)

From equation (i) and (ii) we get

M  =  23 u

So the molar mass of solute is 23 units.

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