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 $\ce{CO2}$ gas on combustion. The percentage of hydrogen in the compound (x) is ______ %. [Nearest Integer] [Given: Molar mass in $\mathrm{g\,mol^{-1}}$ H: 1, C: 12, Br: 80, Ag: 108, O: 16; Compound (x): $\ce{C_xH_yBr_z}$]

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

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 X_{2}

gas on combustion. The percentage of hydrogen in the compound (x) is ______ %. [Nearest Integer]
[Given: Molar mass in

g mo l^{- 1}

H: 1, C: 12, Br: 80, Ag: 108, O: 16; Compound (x):

C X_{x} H X_{y} Br X_{z}

]

Answer: 4

Solution

Combustion of 1.0 g gives 1.32 g

CO X_{2}

, so the carbon present is

\frac{1.32}{44} \times 12 = 0.36

g, i.e. 36%. In the Carius estimation 0.53 g of the compound gives 0.75 g

AgBr

(molar mass 188), so the bromine is

\frac{0.75}{188} \times 80 = 0.319

g from 0.53 g, i.e.

\frac{0.319}{0.53} \times 100 = 60.2%

. As the compound holds only carbon, hydrogen and bromine, the hydrogen is

100 - 36 - 60.2 = 3.8%

, which to the nearest integer is 4%.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

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