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Chemistry · Practical Organic Chemistry

JEE Main 2026 — 28 January, Morning Shift — Question 63

0.53 g 0.53 \mathrm{~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)(\mathrm{x}) gave 1.32 g of CO2\mathrm{CO}_{2} gas on combustion. The percentage of hydrogen in the compound ( x ) is ____\_\_\_\_ %.\%. [Nearest Integer]

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

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

CxHyBrz1 g→CO21.32 g%C=1.32×1244×1×100=36%CxHyBrz0.53 g→AgBr0.75 g%Br=0.75×80188×0.53×100=60.2%%H=100−(36+60.2)%H≃4%\begin{aligned} & \underset{1 \mathrm{~g}}{\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{Br}_{\mathrm{z}}} \rightarrow \begin{array}{c} \mathrm{CO}_{2} 1.32 \mathrm{~g} \end{array} \\& \% \mathrm{C}=\frac{1.32 \times 12}{44 \times 1} \times 100=36 \% \\& \underset{0.53 \mathrm{~g}}{\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{Br}_{\mathrm{z}}} \rightarrow \begin{array}{c} \mathrm{AgBr} 0.75 \mathrm{~g} \end{array} \\& \% \mathrm{Br}=\frac{0.75 \times 80}{188 \times 0.53} \times 100=60.2 \% \\& \% \mathrm{H}=100-(36+60.2) \\& \% \mathrm{H} \simeq 4 \% \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Chemistry
Chapter
Practical Organic Chemistry
Topic
Quantitative organic analysis

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