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

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

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