JEE Main 2024 — Solutions Question with Solution

• Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol^-1
• 1 atm = 760 mm Hg
• Molar mass of water = 18 g.mol^-1

First, we calculate the molality (m) of the solution using the boiling point elevation formula:

$$\Delta T_b=K_b\times m$$

From the problem, we know:

$$2=0.52\times m$$

Solving for m:

$$m=\frac{2}{0.52}\approx 3.846\text{ mol/kg}$$

Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:

$$\frac{\Delta P}{P^0}=\frac{n_{\text{solute}}}{n_{\text{solvent}}}$$

Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$\frac{\Delta P}{P^0}=\frac{m}{1000}\times M_{solvent}$$

Substitute the known values:

$$\Delta P=P^0\times \frac{m}{1000}\times M_{solvent}$$

$$=760\times \frac{3.846}{1000}\times 18$$

$$=52.615\text{ mm Hg}$$

Therefore, the vapor pressure of the solution:

$$P_{solution}=P^0-\Delta P$$

$$=760-52.615$$

$$\approx 707.385\text{ mm Hg}$$

Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.