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

             (i) molar mass of the solute

Answers (1)

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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