The vapour pressure of $30 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of glucose is __________ $\mathrm{mm} ~\mathrm{Hg}$ at $25^{\circ} \mathrm{C}$.

[Given : The density of $30 \%$ (w/v), aqueous solution of glucose is $1.2 \mathrm{~g} \mathrm{~cm}^{-3}$ and vapour pressure of pure water is $24 \mathrm{~mm}~ \mathrm{Hg}$.]

(Molar mass of glucose is $180 \mathrm{~g} \mathrm{~mol}^{-1}$.)

The given solution is a $30 \%$ (w/v) aqueous solution of glucose. This means that $30\ \mathrm{g}$ of glucose is dissolved in $100\ \mathrm{mL}$ of the solution. Since the density of the solution is $1.2\ \mathrm{g/mL}$, the weight of $100\ \mathrm{mL}$ of the solution is: <br/><br/> $$\mathrm{Wt.~of~solution} = 100\ \mathrm{mL} \times 1.2\ \mathrm{g/mL} = 120\ \mathrm{g}$$ <br/><br/> Since the solution is $30 \%$ (w/v), the weight of glucose in $100\ \mathrm{mL}$ of the solution is $30\ \mathrm{g}$. Therefore, the weight of water in $100\ \mathrm{mL}$ of the solution is: <br/><br/> $\mathrm{Wt.~of~water} = 120\ \mathrm{g} - 30\ \mathrm{g} = 90\ \mathrm{g}$ <br/><br/> Now, we can use Raoult's law to calculate the vapour pressure of the solution. Raoult's law states that the partial vapour pressure of a solvent in a solution is proportional to its mole fraction in the solution. For a dilute solution, the mole fraction of the solvent can be approximated as the ratio of the number of moles of solvent to the total number of moles in the solution. <br/><br/> Let us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is: <br/><br/> $$n_{\mathrm{glucose}} = \frac{30\ \mathrm{g}}{180\ \mathrm{g/mol}} = 0.167\ \mathrm{mol}$$ <br/><br/> The number of moles of water in the solution is: <br/><br/> $n_{\mathrm{water}} = \frac{90\ \mathrm{g}}{18\ \mathrm{g/mol}} = 5\ \mathrm{mol}$ <br/><br/> The total number of moles in the solution is: <br/><br/> $$n_{\mathrm{total}} = n_{\mathrm{glucose}} + n_{\mathrm{water}} = 5.167\ \mathrm{mol}$$ <br/><br/> Let $P$ be the vapour pressure of the solution. According to Raoult's law, we have: <br/><br/> $$\frac{P_0 - P}{P} = \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}} = \frac{0.167}{5} = 0.0334$$ <br/><br/> where $P_0$ is the vapour pressure of pure water, which is given as $24\ \mathrm{mmHg}$. <br/><br/> Simplifying the above equation, we get: <br/><br/> $$P = \frac{P_0}{1 + \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}}} = \frac{24\ \mathrm{mmHg}}{1 + \frac{0.167}{5}} = 23.22\ \mathrm{mmHg}$$ <br/><br/> Therefore, the vapour pressure of the $30\%$ (w/v) aqueous solution of glucose at $25^\circ\mathrm{C}$ is $\boxed{23\ \mathrm{mmHg}}$ (rounded off to the nearest integer).