At T ( K ), 100 g of 98 \% H 2 SO 4 ( w / w ) aqueous solution is mixed with 100 g of 49 \% H 2 SO 4 ( w / w ) aqueous solution. What is the mole fraction of H 2 SO 4 in the resultant solution? (Given : Atomic mass H =1 u ; S =32 u ; O =16 u ) (Assume that temperature after mixing remains constant)

Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 22 January, Evening Shift — Question 47

At $\mathrm{T}(\mathrm{K}), 100 \mathrm{~g}$ of $98 \% \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{w} / \mathrm{w})$ aqueous solution is mixed with 100 g of $49 \% \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{w} / \mathrm{w})$ aqueous solution. What is the mole fraction of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the resultant solution? (Given : Atomic mass $\mathrm{H}=1 \mathrm{u} ; \mathrm{S}=32 \mathrm{u} ; \mathrm{O}=16 \mathrm{u}$ ) (Assume that temperature after mixing remains constant)

Option A:
0.9
Option B:
0.1
Option C:
0.337
Correct
Option D:
0.663

Answer: C

Step-by-step solution

Total weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$ $=\left(100 \times \frac{98}{100}\right)+\left(100 \times \frac{49}{100}\right)=147 \mathrm{gm}$ Total weight of $\mathrm{H}_{2} \mathrm{O}=200-147=53 \mathrm{gm}$ Mole fraction of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{\frac{147}{98}}{\left(\frac{147}{98}+\frac{53}{18}\right)}=0.337$

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Concentration Terms and Their Interconversion