Q21 / 75 · JEE Main 2026 (28 January Shift 1) · 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 $(\mathrm{x})$ gave 1.32 g of $\mathrm{CO}_{2}$ gas on combustion. The percentage of hydrogen in the compound $(\mathrm{x})$ is $\_\_\_\_$%.
[Nearest Integer]
[Given: Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{Br}: 80, \mathrm{Ag}: 108, \mathrm{O}: 16$; Compound (x) : $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{Br}_{\mathrm{z}}$]
[Nearest Integer]
[Given: Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{Br}: 80, \mathrm{Ag}: 108, \mathrm{O}: 16$; Compound (x) : $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{Br}_{\mathrm{z}}$]
Correct answer: 4
Solution
From AgBr precipitate:
0.75 g AgBr (M = 188) equals 0.00399 mol Br, so the compound has one Br atom.
From combustion of 1.0 g:
1.32 g CO₂ (M = 44) equals 0.03 mol C.
The molar mass is found from 0.53 g sample:
M = (0.53 × 188)/(0.75 × 1) = 133 g/mol.
From 1.0 g sample producing 0.03 mol CO₂
We have (1.0/133) × x = 0.03, giving x = 4.
Thus the molecular formula is $C_4H_5Br$ with M = 48 + 5 + 80 = 133.
Percentage of hydrogen = (5/133) × 100 = 3.76%, which rounds to 4%.
0.75 g AgBr (M = 188) equals 0.00399 mol Br, so the compound has one Br atom.
From combustion of 1.0 g:
1.32 g CO₂ (M = 44) equals 0.03 mol C.
The molar mass is found from 0.53 g sample:
M = (0.53 × 188)/(0.75 × 1) = 133 g/mol.
From 1.0 g sample producing 0.03 mol CO₂
We have (1.0/133) × x = 0.03, giving x = 4.
Thus the molecular formula is $C_4H_5Br$ with M = 48 + 5 + 80 = 133.
Percentage of hydrogen = (5/133) × 100 = 3.76%, which rounds to 4%.
Hence, the correct answer is 4.