JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of CO $_{2}$ gas on combustion. The percentage of hydrogen in the compound (x) is ______ %. [Nearest Integer]

[Given: Molar mass in g mol $^{- 1}$ H : 1, C : 12, Br : 80, Ag : 108, O : 16; Compound (x) : C $_{x}$ H $_{y}$ Br $_{z}$ ]

Your answer

Solution

Calculate %C from CO $_{2}$ formed in combustion (using 1.0 g of x):

$% C = \frac{12}{44} \times \frac{1.32}{1.0} \times 100 = 36%$

Calculate %Br from AgBr precipitate (using 0.53 g of x):

$% Br = \frac{80}{188} \times \frac{0.75}{0.53} \times 100 \approx 60.2%$

Since x contains only C, H and Br:

$% H = 100 - (36 + 60.2) \approx 3.8% \approx 4%$

$∴$ Answer: $4$ .

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →