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)

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Some Basic Concepts of Chemistry

T (K), 100 g

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)

Answer: (C)

The

100 g

98%

solution contains

98 g

H_{2} S O_{4}

and

2 g

of water; the

100 g

49%

solution contains

49 g

H_{2} S O_{4}

and

51 g

of water. Totals:

H_{2} S O_{4} = 147 g = \frac{147}{98} = 1.5 mol

and water

= 53 g = \frac{53}{18} = 2.944 mol

. The mole fraction of

H_{2} S O_{4}

\frac{1.5}{1.5 + 2.944} = \frac{1.5}{4.444} = 0.337

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option