Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 4 April, Shift 1 — Question 81

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at $25^{\circ} \mathrm{C}$ . The solution showed a boiling point elevation by $2^{\circ} \mathrm{C}$ . Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ______ mm of 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 atm pressure $=760 \mathrm{~mm}$ of Hg , molar mass of water $\left.=18 \mathrm{~g} \mathrm{~mol}^{-1}\right]$

Answer: 711

Numerical answer — enter this value.

Step-by-step solution

Given: Mass of solute $= 2.5\,\mathrm{g}$ ; mass of water $= 100\,\mathrm{g}=0.1\,\mathrm{kg}$ ; $\Delta T_b = 2\,\mathrm{K}$ ; $K_b(\text{water})=0.52\,\mathrm{K\,kg\,mol^{-1}}$ ; $P^0(\text{water})=760\,\mathrm{mm\,Hg}$ ; molar mass of water $=18\,\mathrm{g\,mol^{-1}}$

Molality, $m=\dfrac{\Delta T_b}{K_b}=\dfrac{2}{0.52}=3.846\,\mathrm{mol\,kg^{-1}}$

Moles of solute, $n_{\text{solute}}=3.846\times0.1=0.3846\,\mathrm{mol}$

Moles of water, $n_{\text{water}}=\dfrac{100}{18}=5.56\,\mathrm{mol}$

Mole fraction of water, $X_{\text{water}}=\dfrac{5.56}{5.56+0.3846}=0.935$

Vapour pressure of solution, $P=X_{\text{water}}P^0=0.935\times760=710.6\approx711\,\mathrm{mm\,Hg}$ .

Thus, vapour pressure of the solution $=711\,\mathrm{mm\,Hg}$ .

Answer key and solution verified before publishing.

Practise Solutions and Colligative Properties

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)