QJEEM24Ch443f: JEE Main 2024 Chemistry - Class 12 JEEM24Ch443f Solution: of a non-volatile, non-electrolyte is dissolved in of water at…

• 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.