JEE Main 2026 — Jan 22, Shift 2 — Chemistry solution

JEE Main2026Jan 22, Shift 2Chemistry

At $T (K)$ , 100 g of 98% $H_{2} S O_{4}$ (w/w) aqueous solution is mixed with 100 g of 49% $H_{2} S O_{4}$ (w/w) aqueous solution. What is the mole fraction of $H_{2} S O_{4}$ in the resultant solution ?

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

(Assume that temperature after mixing remains constant)

A
0.9
B
0.1
C
0.337
D
0.663

Solution

Mass of $H_{2} S O_{4}$ in the mixture: $100 \times 0.98 + 100 \times 0.49 = 147$ g.

Mass of water: $200 - 147 = 53$ g.

Moles: $n_{H_{2} S O_{4}} = \frac{147}{98} = 1.5$ , $n_{H_{2} O} = \frac{53}{18} \approx 2.944$ .

Mole fraction of $H_{2} S O_{4} = \frac{1.5}{1.5 + 2.944} \approx 0.337$ .

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