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# A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa 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: molar mass of the solute

$\begin{array}{1 1}23g mol^{-1}\\43g mol^{-1}\\33g mol^{-1}\\53g mol^{-1}\end{array}$

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## 1 Answer

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Suppose the molar mass of the solute =$M\;gmol^{-1}$
$n_2$ (solute) =$\large\frac{30}{M}$ moles
$n_1$ (solvent,$H_2O$)=$\large\frac{90g}{18g mol^{-1}}$=5 moles
$\large\frac{P^0-P_s}{p^0}=\large\frac{\Large\frac{30}{M}}{5+\Large\frac{30}{M}}$
$1-\large\frac{2.8}{P^0}=\large\frac{\Large\frac{30}{M}}{5+\Large\frac{30}{M}}$
$\large\frac{2.8}{P^0}$$=1-\large\frac{\Large\frac{30}{M}}{5+\Large\frac{30}{M}} \large\frac{2.8}{P^0}=\large\frac{5+\Large\frac{30}{M}-\frac{30}{M}}{5+\Large\frac{30}{M}} \large\frac{P^0}{2.8}=\large\frac{5+\Large\frac{30}{M}}{5}$$=1+\large\frac{6}{M}$-----(i)
Adding 18 g of water
$n_1=$6 moles
$\large\frac{P^0-2.9}{P^0}=\large\frac{\Large\frac{30}{M}}{6+\Large\frac{30}{M}}$
$1-\large\frac{2.9}{P^0}=\large\frac{\Large\frac{30}{M}}{6+\Large\frac{30}{M}}$
$\large\frac{2.9}{P^0}$$=1-\large\frac{\Large\frac{30}{M}}{6+\Large\frac{30}{M}} \large\frac{P^0}{2.9}=\large\frac{6+\Large\frac{30}{M}}{6}$$=1+\large\frac{5}{M}$-----(ii)
Dividing equation (i) by equation (ii) we get
$\large\frac{2.9}{2.8}=\large\frac{1+\Large\frac{6}{M}}{1+\Large\frac{5}{M}}$
(or) $2.9(1+\large\frac{5}{M})$$=2.8(1+\large\frac{6}{M}) 2.9+\large\frac{14.5}{M}$$=2.8+\large\frac{16.8}{M}$
(or) $\large\frac{2.3}{M}$$=0.1$
$M=23gmol^{-1}$
answered Aug 5, 2014

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