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

JEE Main2026Jan 22, Shift 2Chemistry

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL, 1M $H_{2} S O_{4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is :

A
21
B
0.42
C
42
D
0.21

Solution

Equivalents of $H_{2} S O_{4}$ = equivalents of ammonia. With $H_{2} S O_{4}$ as a diprotic acid:

Moles of $N H_{3} = 2 \times 15 \times 1 0^{- 3} \times 1 = 0.030$ mol.

This equals the moles of nitrogen in the sample. Mass of N $= 0.030 \times 14 = 0.42$ g.

% N $= \frac{0.42}{1} \times 100 = 42%$ .

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