A solution contains 20 g glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ and 34.2 g of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be

Amount of glucose, $n_2=\frac{m_2}{M_2}=\frac{20\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}=\frac{1}{9}\mathrm{mol}$; Amount of sucrose, $n_3=\frac{m_3}{M_3}=\frac{34.2\mathrm{g}}{342\mathrm{g}{\mathrm{mol}}^{-1}}=\frac{1}{10}\mathrm{mol}$

Amount of water, $n_1=\frac{m_1}{M_1}=\frac{108\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=6\mathrm{mol}$

Mole fraction of solutes in solution,

$x=\frac{n_2+n_3}{n_1+n_2+n_3}=\frac{\left[\right(1/9)+(1/10\left)\right]\mathrm{mol}}{6\mathrm{mol}+\lfloor (1/9)+(1/10)\mathrm{mol}]}=\frac{0.211\mathrm{mol}}{6.211\mathrm{mol}}=0.034$

Vapour pressure of solution. $p_1=x_2{p}_1^{\ast }=\left(0.034\right)\left(35\mathrm{mmHg}\right)=1.19\mathrm{mmHg}$