Q10 / 75 · JEE Main 2026 (22 January Shift 2) · Chemistry
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)
(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)
- A0.663
- B0.9
- Correct answerC0.337
- D0.1
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.
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.
Hence, the correct answer is option: 'C'.