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JEE Main Chemistry Solutions 2026 JEE Main 2026 (22 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

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)

Options

  1. A. 0.663
  2. B. 0.9
  3. C. 0.337
  4. D. 0.1

Answer

C. 0.337

Step-by-step solution

Molar mass of H₂SO₄ = 98 g/mol. Solution 1: 100 g at 98% gives 98 g H₂SO₄ = 1 mol; water = 2 g = 0.111 mol. Solution 2: 100 g at 49% gives 49 g H₂SO₄ = 0.5 mol; water = 51 g = 2.833 mol. Mixed solution: Total H₂SO₄ = 1 + 0.5 = 1.5 mol. Total water = 0.111 + 2.833 = 2.944 mol. Mole fraction of H₂SO₄ = 1.5/(1.5 + 2.944) = 1.5/4.444 = 0.337.

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