QJEEM23Chbf31: JEE Main 2023 Chemistry - Class 12 JEEM23Chbf31 Solution: The vapour pressure of aqueous solution of glucose is…

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:

$\mathrm{Wt.~of~solution} = 100\ \mathrm{mL} \times 1.2\ \mathrm{g/mL} = 120\ \mathrm{g}$

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:

$\mathrm{Wt.~of~water} = 120\ \mathrm{g} - 30\ \mathrm{g} = 90\ \mathrm{g}$

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.

Let us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is:

$n_{\mathrm{glucose}} = \frac{30\ \mathrm{g}}{180\ \mathrm{g/mol}} = 0.167\ \mathrm{mol}$

The number of moles of water in the solution is:

$n_{\mathrm{water}} = \frac{90\ \mathrm{g}}{18\ \mathrm{g/mol}} = 5\ \mathrm{mol}$

The total number of moles in the solution is:

$n_{\mathrm{total}} = n_{\mathrm{glucose}} + n_{\mathrm{water}} = 5.167\ \mathrm{mol}$

Let $P$ be the vapour pressure of the solution. According to Raoult's law, we have:

$\frac{P_0 - P}{P} = \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}} = \frac{0.167}{5} = 0.0334$

where $P_0$ is the vapour pressure of pure water, which is given as $24\ \mathrm{mmHg}$ .

Simplifying the above equation, we get:

$P = \frac{P_0}{1 + \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}}} = \frac{24\ \mathrm{mmHg}}{1 + \frac{0.167}{5}} = 23.22\ \mathrm{mmHg}$

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).