JEE Chemistry 2023 Question with Solution

Given: $30\%\ (w/v)$ glucose solution, density $= 1.2\ \mathrm{g\,cm^{-3}}$, vapour pressure of pure water $= 24\ \mathrm{mm\ Hg}$, molar mass of glucose $= 180\ \mathrm{g\,mol^{-1}}$.

Find: Vapour pressure of the solution.

Interpret $30\%\ (w/v)$ solution: $30\ \mathrm{g}$ of glucose is present in $100\ \mathrm{mL}$ of solution.

Mass of solution:

$$\text{Mass of solution} = 1.2 \times 100 = 120\ \mathrm{g}$$

Mass of water:

$$\text{Mass of water} = 120 - 30 = 90\ \mathrm{g}$$

Moles of glucose and water:

$$n_{\text{glucose}} = \frac{30}{180} = 0.167\ \mathrm{mol}$$ $$n_{\text{water}} = \frac{90}{18} = 5\ \mathrm{mol}$$

Mole fraction of water:

$$X_{\text{water}} = \frac{5}{5 + 0.167} \approx 0.967$$

Apply Raoult's law:

$$P_{\text{solution}} = X_{\text{water}} \times P^\circ_{\text{water}}$$ $$P_{\text{solution}} = 0.967 \times 24 \approx 23.2\ \mathrm{mm\ Hg}$$

Nearest integer value:

$$P_{\text{solution}} \approx 23\ \mathrm{mm\ Hg}$$

Therefore, the vapour pressure of the solution is $23\ \mathrm{mm\ Hg}$.

Given: The solute is non-volatile glucose.

Find: Vapour pressure of the solution.

Use Raoult's law directly after finding the mole fraction of the solvent. The key idea is that only water contributes to vapour pressure, so

$$P = X_{\text{solvent}} P^\circ$$

Compute the mass of water from density and total volume first, then convert both solute and solvent to moles. After obtaining $X_{\text{water}} \approx 0.967$, multiply by $24\ \mathrm{mm\ Hg}$ to get approximately $23.2\ \mathrm{mm\ Hg}$, which gives $23$ as the required integer value.