ChemistryJEE Main 22 Jan 2026, Shift 2·Q1·MCQ·Easy–moderate

Some Basic Concepts · Mole fraction of mixed solutions

At T(K), 100 g of 98% $\ce{H2SO4}$ (w/w) aqueous solution is mixed with 100 g of 49% $\ce{H2SO4}$ (w/w) aqueous solution. What is the mole fraction of $\ce{H2SO4}$ in the resultant solution ?

(Given : Atomic mass H = 1 u ; S = 32 u; O = 16 u)

(Assume that temperature after mixing remains constant)

(1)0.9
(2)0.1
(3)0.337
(4)0.663

We corrected a printing error in the original paper

Answer corrected; the source key is (4) — 0.663 is the mole fraction of water; $\ce{H2SO4}$ = 147/98 = 1.5 mol, water = 53/18 = 2.94 mol, so x( $\ce{H2SO4}$ ) = 0.337.

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