At T K, 100 g of 98 H2SO4 (w/w) aqueous solution is mixed with 10… — Solution | JEE Chemistry

MCQEasyJEE 2026Concentration Terms (Molarity, Molality, etc.)

At $T$ K, $100 \, g$ of $98\%$ $H_2SO_4$ (w/w) aqueous solution is mixed with $100 \, g$ of $49\%$ $H_2SO_4$ (w/w) aqueous solution. What is the mole fraction of $H_2SO_4$ in the resultant solution? (Given: Atomic mass $H = 1 \, u,\; S = 32 \, u,\; O = 16 \, u$ . Assume that temperature after mixing remains constant.)

A
$0.337$
B
$0.1$
C
$0.9$
D
$0.663$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $100 \, g$ of $98\%$ $H_2SO_4$ solution is mixed with $100 \, g$ of $49\%$ $H_2SO_4$ solution.

Find: Mole fraction of $H_2SO_4$ in the resultant solution.

Step 1: Calculate masses of $H_2SO_4$ and water.

From $100 \, \text{g}$ of $98\%$ solution:

\text{Mass of } H_2SO_4 = 98 \, \text{g}, \quad \text{Mass of } H_2O = 2 \, \text{g}

From $100 \, \text{g}$ of $49\%$ solution:

\text{Mass of } H_2SO_4 = 49 \, \text{g}, \quad \text{Mass of } H_2O = 51 \, \text{g}

Step 2: Total masses after mixing.

\text{Total } H_2SO_4 = 98 + 49 = 147 \, \text{g}

\text{Total } H_2O = 2 + 51 = 53 \, \text{g}

Step 3: Convert masses into moles. Molar mass of $H_2SO_4 = 98 \, \text{g mol}^{-1}$ ,

n(H_2SO_4) = \frac{147}{98} = 1.5 \, \text{mol}

Molar mass of $H_2O = 18 \, \text{g mol}^{-1}$ ,

n(H_2O) = \frac{53}{18} \approx 2.94 \, \text{mol}

Step 4: Calculate mole fraction of $H_2SO_4$ .

X_{H_2SO_4} = \frac{1.5}{1.5 + 2.94} = \frac{1.5}{4.44} \approx 0.337

Therefore, the mole fraction of $H_2SO_4$ is $0.337$ . The correct option is A.

Common mistakes

Using percentage values directly as mole fractions is incorrect because $98\%$ and $49\%$ are mass percentages, not mole fractions. First convert each percentage into actual masses of $H_2SO_4$ and $H_2O$ .
Adding solution percentages to get the final concentration is wrong because mixing must be handled through total masses of solute and solvent. Compute total mass of $H_2SO_4$ and total mass of $H_2O$ after mixing.
Forgetting to include water in the denominator of mole fraction gives an incorrect result. Mole fraction of $H_2SO_4$ is $\frac{n(H_2SO_4)}{n(H_2SO_4)+n(H_2O)}$ , not just moles of acid alone.

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Step-by-step solution

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