Answer is : 100
Organic compound \(\xrightarrow {\text{combustion}} \underset{0.9gm}{H_2O}\)
\(\therefore\) mole of \(\mathrm{H}_{2} \mathrm{O}=\frac{0.9}{18}=0.05\) mole
\(\therefore\) mole of H in \(\mathrm{H}_{2} \mathrm{O}=0.05 \times 2=0.1\) mole
= mole of H in 0.01 mole Organic compound
\(\therefore \ \mathrm{wt}\) of H atom in 0.01 mole compound \(=0.1 \times 1 = 0.1\ gm\)
\(\therefore\) wt of H atom in one mole compound \(=\frac{0.1}{0.01}=10 \mathrm{gm}\)
\(\because\) wt. \(\%\) of \(\mathrm{H}=\frac{\text { wt. of } \mathrm{H} \text { in one mole compound }}{\text { Molar mass of compound }} \times 1\)
\(10=\frac{10}{M} \times 100\)
\(\therefore \mathrm{M}=100\)
