JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

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 $X =$ ______ $\times 1 0^{- 3}$ g. (Nearest integer)

(Molar mass of C, H and O are $12$ , $1$ and $16$ g mol $^{- 1}$ respectively.)

A
$273$
B
$27$
C
$2730$
D
$227$

Solution

Mass of carbon in the sample $= \frac{12}{44} \times 0.25$ g.

Given carbon percentage $= 25%$ , so $\frac{( 12/44 ) ( 0.25 )}{X} \times 100 = 25$ .

Solving: $X = \frac{12 \times 0.25 \times 100}{44 \times 25} = \frac{12}{44} = 0.2727$ g $\approx 273 \times 1 0^{- 3}$ g.

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