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Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 2 April, Evening Shift — Question 18

When 1 g each of compounds ABA B and AB2A B_{2} are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is _____\_\_\_\_\_ ×10−1\times 10^{-1} (Nearest integer) (Given: Molal boiling point elevation constant is 0.5 Kkgmol−1\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} )

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

For AB ΔTb=2.7\Delta \mathrm{T}_{\mathrm{b}}=2.7

Kb=0.5 K kg mol−1\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

m=ΔTbKb=5.4 m\mathrm{m}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}}}=5.4 \mathrm{~m}

For AB2A B_{2}

ΔTb=1.5 K\Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K}

m=1.50.5=5.4 m\mathrm{m}=\frac{1.5}{0.5}=5.4 \mathrm{~m}

Moles of AB=0.081 molA B=0.081 \mathrm{~mol}

Moles of AB2A B_{2} =0.045 mol=0.045 \mathrm{~mol}

Molar mass of AB=12.35 g/mol\mathrm{AB}=12.35 \mathrm{~g} / \mathrm{mol}

Molar mass of AB2=22.22 g/mol\mathrm{AB}_{2}=22.22 \mathrm{~g} / \mathrm{mol}

Let atomic mass of A=a,B=b\mathrm{A}=\mathrm{a}, \mathrm{B}=\mathrm{b}

a+b=12.35a+b=12.35

a+2b=22.22a+2 b=22.22

On solving a=2.48\mathrm{a}=2.48

=24.8×10−1=24.8 \times 10^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Solid in Liquid Solutions (Colligative Properties)