0.01 mol of an organic compound (X) containi | vedclass

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(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$. Moles of $H$ atoms $= 0.05 \ mol 	imes 2 = 0.1 \ mol$. Mass of $H$ a

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$100$

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(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$. Moles of $H$ atoms $= 0.05 \ mol 	imes 2 = 0.1 \ mol$. Mass of $H$ atoms in $0.01 \ mol$ of $(X) = 0.1 \ mol 	imes 1 \ g \ mol^{-1} = 0.1 \ g$. Mass of $H$ atoms in $1 \ mol$ of $(X) = \frac{0.1 \ g}{0.01 \ mol} = 10 \ g \ mol^{-1}$. Given that the compound contains $10\%$ hydrogen by mass,we have: $10\% = \frac{	ext{Mass of } H 	ext{ in } 1 \ mol 	ext{ of } (X)}{	ext{Molar mass of } (X)} 	imes 100$. $10 = \frac{10}{M} 	imes 100$. $M = 100 \ g \ mol^{-1}$.

$0.01 \ mol$ of an organic compound $(X)$ containing $10\%$ hydrogen,on complete combustion produced $0.9 \ g \ H_2O$. Molar mass of $(X)$ is $.......... \ g \ mol^{-1}$.

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