The molecular formula of an alkane having a w | vedclass

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(C) At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$ volume.Given: Weight = $11 \ g$,Volume = $5.6 \ L$.Number of moles = $\frac{	ext{Given Volume}

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$C_3H_8$

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(C) At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$ volume.Given: Weight = $11 \ g$,Volume = $5.6 \ L$.Number of moles = $\frac{	ext{Given Volume}}{	ext{Molar Volume at STP}} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$.Molar mass $(M)$ = $\frac{	ext{Weight}}{	ext{Moles}} = \frac{11 \ g}{0.25 \ mol} = 44 \ g/mol$.For an alkane with formula $C_nH_{2n+2}$,the molar mass is $12n + 1(2n+2) = 14n + 2$.Setting $14n + 2 = 44$,we get $14n = 42$,so $n = 3$.Therefore,the molecular formula is $C_3H_8$.

The molecular formula of an alkane having a weight of $11 \ g$ and a volume of $5.6 \ L$ at $STP$ is.......

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