JEE MAIN - Chemistry (2024 - 4th April Morning Shift - No. 21)
$$2.5 \mathrm{~g}$$ of a non-volatile, non-electrolyte is dissolved in $$100 \mathrm{~g}$$ of water at $$25^{\circ} \mathrm{C}$$. The solution showed a boiling point elevation by $$2^{\circ} \mathrm{C}$$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ $$\mathrm{mm}$$ of $$\mathrm{Hg}$$ (nearest integer)
[Given : Molal boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}$, $$1 \mathrm{~atm}$$ pressure $$=760 \mathrm{~mm}$$ of $$\mathrm{Hg}$$, molar mass of water $$=18 \mathrm{~g} \mathrm{~mol}^{-1}]$$
Explanation
- 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.
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