The vapour pressure of water at $20 °C$ is 17.5 mmHg. If 18 g of glucose $(C_6{H}_{12}{O}_6 )$ is added to 178.2 g of water at $20 °C$, the vapour of the resulting solution will be

Amount of glucose, $n_2=\frac{m_2}{M_2}=\frac{18\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}=0.1\mathrm{mol}$

Amount of water, $n_1=\frac{m_1}{M_1}=\frac{178.2}{18\mathrm{g}{\mathrm{mol}}^{-1}}=9.9\mathrm{mol}$

Mole fraction of water, $x_1=\frac{n_1}{n_1+n_2}=\frac{9.9\mathrm{g}}{(9.9+0.1)\mathrm{mol}}=0.99$

Vapour pressure of solution, $p=x_1{p}_1^{\ast }=\left(0.99\right)\left(17.5\mathrm{mmHg}\right)=17.325\mathrm{mmHg}.$