Chemistry · Solutions and Colligative Properties

JEE Advanced 2024 — Paper 2 — Question 31

Vessel-1 contains w2g\mathbf{w}_{\mathbf{2}} \mathrm{g} of a non-volatile solute X\mathbf{X} dissolved in w1g\mathbf{w}_{\mathbf{1}} \mathrm{g} of water. Vessel-2 contains w2\mathbf{w}_{\mathbf{2}} gg of another non-volatile solute Y\mathbf{Y} dissolved in w1 g\mathbf{w}_{1} \mathrm{~g} of water. Both the vessels are at the same temperature and pressure. The molar mass of X\mathbf{X} is 80%80 \% of that of Y\mathbf{Y}. The van't Hoff factor for X\mathbf{X} is 1.2 times of that of Y\mathbf{Y} for their respective concentrations.

The elevation of boiling point for solution in Vessel-1 is _____\_\_\_\_\_ % of the solution in Vessel-2.

Answer: 150

Numerical answer — enter this value.

Step-by-step solution

(ΔTb)1=i1kf×W2Mx\left(\Delta T_{b}\right)_{1}=i_{1} k_{\mathrm{f}} \times \frac{W_{2}}{M_{x}}

(W11000)‾\overline{\left(\frac{W_{1}}{1000}\right)}

(ΔTb)2=i2kf×W2My(W1)\left(\Delta T_{b}\right)_{2}=i_{2} \mathrm{k}_{\mathrm{f}} \times \frac{\mathrm{W}_{2}}{\frac{\mathrm{M}_{\mathrm{y}}}{\left(W_{1}\right)}}

(W11000)‾\overline{\left(\frac{W_{1}}{1000}\right)}

(ΔTb)1(ΔTb)2=i1i2×MyMx=1.2×10.8=1.5\frac{\left(\Delta T_{b}\right)_{1}}{\left(\Delta T_{b}\right)_{2}}=\frac{i_{1}}{i_{2}} \times \frac{M_{y}}{M_{x}}=1.2 \times \frac{1}{0.8}=1.5 (ΔTb)1=1.5(Δ Tb)2\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{1}=1.5\left(\Delta \mathrm{~T}_{\mathrm{b}}\right)_{2}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Solid in Liquid Solutions (Colligative Properties)