JEE Main 202624 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 24 January
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- General Organic Chemistry
JEE Main 2026 · 24 January · Evening Shift · Chemistry · General Organic Chemistry
0.25 g of an organic compound " A " containing carbon, hydrogen and...
0.25 g of an organic compound " A " containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is %. (Nearest integer)
(Given: molar mass in )
(Given: molar mass in )
Correct answer
0
Step-by-step solution
In combustion analysis: CaCl₂ tube absorbs H₂O and potash tube absorbs CO₂.
Mass increase of CaCl₂ = 0.15 g gives moles of H₂O = 0.15/18 = 0.00833 mol, so moles of H = 0.01667 mol and mass of H = 0.01667 g.
Mass increase of potash = 0.1837 g gives moles of CO₂ = 0.1837/44 = 0.00417 mol, so moles of C = 0.00417 mol and mass of C = 0.05 g.
Mass of O = 0.25 - 0.05 - 0.01667 = 0.18333 g.
Percentage of O = (0.18333/0.25) × 100 = 73.33% ≈ 73% (nearest integer).
Mass increase of CaCl₂ = 0.15 g gives moles of H₂O = 0.15/18 = 0.00833 mol, so moles of H = 0.01667 mol and mass of H = 0.01667 g.
Mass increase of potash = 0.1837 g gives moles of CO₂ = 0.1837/44 = 0.00417 mol, so moles of C = 0.00417 mol and mass of C = 0.05 g.
Mass of O = 0.25 - 0.05 - 0.01667 = 0.18333 g.
Percentage of O = (0.18333/0.25) × 100 = 73.33% ≈ 73% (nearest integer).