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# Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

$\begin{array}{1 1}17.44mm\\18.44mm\\19.44mm\\15.44mm\end{array}$

Given
$P^0=17.535mm$
$W_2=25g$
$W_1=450g$
$M_2=C_6H_{12}O_6=72+12+96$
$\Rightarrow 180g mol^{-1}$
For solvent $(H_2O)$
$M_1=18gmol^{-1}$
Applying Raoult's law
$\large\frac{P^0-P_s}{P^0}=\frac{n_2}{n_1+n_2}$
(or) $\large\frac{P^0-P_s}{P^0}=\frac{n_2}{n_1}$
$\large\frac{P^0-P_s}{P^0}=\frac{W_2}{M_2}\times \frac{M_1}{W_1}$
$1-\large\frac{P_s}{P^0}=\frac{25\times 18}{M_2 \times 450}$
$1-\large\frac{P_s}{P^0}=\frac{25\times 18}{180 \times 450}$
$1-\large\frac{P_s}{17.535}=\frac{25\times 18}{180 \times 450}$
$\large\frac{17.535}{P_s}=$$1+\large\frac{25}{4500} \large\frac{17.535}{P_s}=$$\large\frac{4525}{4500}$
$P_s=\large\frac{4500\times 17.535}{4525}$
$\;\;\;\;\;=17.44$mm