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  4. Relative decrease in V.P. of an aqueous solution of NaCl is 0.167. Number of moles of NaCl presentin 180 g of H2O is

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Q. Relative decrease in V.P. of an aqueous solution of NaCl is 0.167. Number of moles of NaCl presentin 180 g of $H_2O$ is

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Solution:

$\frac{\Delta_p}{p} = x_B$
$x_B$ = 0.167
No. of moles of water = $\frac{180}{18}$ = 10 mol
Let no. of moles of NaCl = $n$
Van't Hoff factor i for NaCl = 2
$\therefore $ $x_B = \frac{2n}{2n + 10}$
$\frac{2n}{2n+10} = 0.167 \, \Rightarrow \, \frac{n}{n+5} $ = 0.167
$n = 0.167 \, n + 0.845 $
$0.833 \, n = 0.835 \, \Rightarrow \, n= \frac{0.835}{0.833} = 1 $