JEE Main 2026 — Solutions Question with Solution
JEE Main 2026 (22 January Shift 2)
Question
At of aqueous solution is mixed with 100 g of aqueous solution. What is the mole fraction of in the resultant solution?
(Given : Atomic mass ).
(Assume that temperature after mixing remains constant)
(Given : Atomic mass ).
(Assume that temperature after mixing remains constant)
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.337
Step-by-step explanation
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.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Solutions chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2026, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.