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Chemistry · Practical Organic Chemistry

JEE Main 2026 — 2 April, Evening Shift — Question 57

Complete combustion of X g of an organic compound gave 0.25 g of CO2\mathrm{CO}_{2} and 0.12 g of H2O\mathrm{H}_{2} \mathrm{O}. If the % of carbon is 25%25 \% and of hydrogen is 4.89%4.89 \%, then X=\mathrm{X}= ____\_\_\_\_ ×10−3 g\times 10^{-3} \mathrm{~g}. (Nearest integer) (Molar mass of C,H\mathrm{C}, \mathrm{H} and O are 12, 1 and 16 g mol−116 \mathrm{~g} \mathrm{~mol}^{-1} respectively.)

  1. Option A:

    273

    Correct
  2. Option B:

    27

  3. Option C:

    2730

  4. Option D:

    227

Answer: A

Step-by-step solution

Mass of carbon =0.2544×12=\frac{0.25}{44} \times 12 Mass% of carbon = Mass of carbon x×100x=0.25×1244×10025=\frac{\text { Mass of carbon }}{\mathrm{x}} \times 100 \mathrm{x}=\frac{\frac{0.25 \times 12}{44} \times 100}{25} x=1244\mathrm{x}=\frac{12}{44} x=0.2727=272.7×10−3\mathrm{x}=0.2727=272.7 \times 10^{-3} ≃273×10−3\simeq 273 \times 10^{-3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Chemistry
Chapter
Practical Organic Chemistry
Topic
Quantitative organic analysis

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