0.092 g of a compound with the molecular for | vedclass

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(C) The molar mass of $C_3H_8O_3$ is $(3 	imes 12) + (8 	imes 1) + (3 	imes 16) = 36 + 8 + 48 = 92 \ g/mol$.Number of moles of the compound $= \fra

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$3$

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(C) The molar mass of $C_3H_8O_3$ is $(3 	imes 12) + (8 	imes 1) + (3 	imes 16) = 36 + 8 + 48 = 92 \ g/mol$.Number of moles of the compound $= \frac{0.092 \ g}{92 \ g/mol} = 0.001 \ mol$.Reaction with $CH_3MgI$ produces $CH_4$ gas corresponding to the number of active hydrogen atoms $(x)$: $Compound + x CH_3MgI ightarrow x CH_4$.Number of moles of $CH_4$ produced at $STP = \frac{67.00 \ mL}{22400 \ mL/mol} \approx 0.00299 \ mol \approx 0.003 \ mol$.Since $1 \ mol$ of compound gives $x \ mol$ of $CH_4$,we have $0.001 	imes x = 0.003$.Therefore,$x = 3$. The number of active hydrogen atoms is $3$.

$0.092 \ g$ of a compound with the molecular formula $C_3H_8O_3$ on reaction with an excess of $CH_3MgI$ gives $67.00 \ mL$ of methane at $STP$. The number of active hydrogen atoms present in a molecule of the compound is

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