JEE Main 202628 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 28 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- General Organic Chemistry
JEE Main 2026 · 28 January · Morning Shift · Chemistry · General Organic Chemistry
0.53 g of an organic compound (x) when heated with excess of nitric...
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 gave 1.32 g of gas on combustion. The percentage of hydrogen in the compound is %.
[Nearest Integer]
[Given: Molar mass in ; Compound (x) : ]
[Nearest Integer]
[Given: Molar mass in ; Compound (x) : ]
Correct answer
0
Step-by-step 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 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 with M = 48 + 5 + 80 = 133.
Percentage of hydrogen = (5/133) × 100 = 3.76%, which rounds to 4%.