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# An aqueous solution of 2 % non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent.What is the molar mass of the solute?

Q2.15  An aqueous solution of  $2\%$ non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

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It is given that $2\%$ of aq. solution. This means 2 g of non-volatile solute in 98 g of H2O.

Also the vapour of water at normal boiling point = 1.013 bar.

Using Raoult's law :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{w_2.M_1}{w_1.M_2}$

So we get :

$M_2 = \frac{2\times18\times1.013}{0.009\times98} = 41.35\ g\ mol^{-1}$

Thus the molar mass of non-volatile solute is 41.35 unit.

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