QJEEM25Ch2f83: JEE Main 2025 Chemistry - Class 11 JEEM25Ch2f83 Solution: 0.01 mole of an organic compound containing hydrogen, on…

Organic compound $\xrightarrow{\text { combustion }} \underset{\substack{0.9 \mathrm{gm}}}{\mathrm{H}_2 \mathrm{O}}$

$\begin{aligned}& \therefore \text { mole of } \mathrm{H}_2 \mathrm{O}=\frac{0.9}{18}=0.05 \text { mole } \\& \begin{aligned}\therefore \text { mole of } \mathrm{H} \text { in } \mathrm{H}_2 \mathrm{O} & =0.05 \times 2=0.1 \text { mole } \\& =\text { mole of } \mathrm{H} \text { in } 0.01 \text { mole } \\& \text { Organic compound }\end{aligned}\end{aligned}$

$\begin{aligned}&\begin{aligned}\therefore \text { wt of } \mathrm{H} \text { atom in } 0.01 \text { mole compound } & =0.1 \times 1 \\& =0.1 \mathrm{gm}\end{aligned}\\&\therefore \text { wt of } \mathrm{H} \text { atom in one mole compound }\\&\begin{aligned}& =\frac{0.1}{0.01}=10 \mathrm{gm} \\& \because \text { wt. } \% \text { of } H=\frac{\text { wt. of } H \text { in one mole compound }}{\text { Molar mass of compound }} \times 1 \\& 10=\frac{10}{M} \times 100 \\& \therefore M=100\end{aligned}\end{aligned}$