Q12 / 75 · JEE Main 2026 (02 April Shift 2) · Chemistry

Complete combustion of $X$ g of an organic compound gave $0.25$ g of $CO_2$ and $0.12$ g of $H_2 O$. If the % of carbon is $25\%$ and of hydrogen is $4.89\%$, then $X = $ _____ $\times 10^{-3}$ g. (Nearest integer) (Molar mass of C, H and O are $12, 1$ and $16$ g mol$^{-1}$ respectively.)

  • Correct answer
    A
    $273$
  • B
    $27$
  • C
    $2730$
  • D
    $227$

Solution

Mass of carbon in the compound is calculated from the mass of $CO_2$ produced:

Mass of C $= \dfrac{12}{44} \times \text{mass of } CO_2 = \dfrac{12}{44} \times 0.25 = \dfrac{3}{44}$ g

Given that the percentage of carbon in the compound is $25\%$:

$\% \text{ of C} = \dfrac{\text{Mass of C}}{\text{Total mass } X} \times 100$

$25 = \dfrac{\dfrac{3}{44}}{X} \times 100$

$X = \dfrac{3}{44} \times \dfrac{100}{25} = \dfrac{3}{44} \times 4 = \dfrac{3}{11}$ g

$X \approx 0.2727$ g

Verifying this using the percentage of hydrogen:

Mass of H $= \dfrac{2}{18} \times \text{mass of } H_2O = \dfrac{2}{18} \times 0.12 = \dfrac{1}{75}$ g

$\% \text{ of H} = \dfrac{\dfrac{1}{75}}{X} \times 100 = 4.89$

$X = \dfrac{100}{75 \times 4.89} \approx 0.2727$ g

Thus, $X = 272.7 \times 10^{-3}$ g.

Rounding to the nearest integer, the required value is $273$.

Answer: $273$

Hence, the correct answer is option: 'A'.

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