Let the molar mass of the compound be $M\,\text{g mol}^{-1}$.

Mass of $0.01$ mole of compound $\text{mass of }X = 0.01M\,\text{g}$

Given that the compound contains $10\%$ hydrogen by mass So, mass of hydrogen in $0.01$ mole of $X$ is $\text{mass of H} = 10\%\text{ of }0.01M = 0.10\times 0.01M = 0.001M\,\text{g}$

From combustion data Complete combustion produces $0.9\,\text{g}$ of water.

In water, the mass fraction of hydrogen is $\frac{2}{18} = \frac{1}{9}$

Therefore, mass of hydrogen present in $0.9\,\text{g}$ of water is $0.9\times \frac{1}{9} = 0.1\,\text{g}$

Equate hydrogen from the compound and hydrogen in water $0.001M = 0.1$

Solve for $M$ $M = \frac{0.1}{0.001} = 100$

Final answer $\boxed{100\,\text{g mol}^{-1}}$

Comparison with stored answer

Stored correct answer = $100$

Our derived answer matches the stored answer.