A gaseous mixture contains CH_4 and C_2H_6 in | vedclass

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(C) At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.Therefore,$2.24 \ L$ of the mixture corresponds to $\frac{2.24}{22.4} = 0.1 \ mol$ of the mix

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$2.3$

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(C) At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.Therefore,$2.24 \ L$ of the mixture corresponds to $\frac{2.24}{22.4} = 0.1 \ mol$ of the mixture.Since the mixture is equimolecular,it contains equal moles of $CH_4$ and $C_2H_6$.Moles of $CH_4 = 0.05 \ mol$ and moles of $C_2H_6 = 0.05 \ mol$.Mass of $CH_4 = 0.05 \ mol 	imes 16 \ g/mol = 0.8 \ g$.Mass of $C_2H_6 = 0.05 \ mol 	imes 30 \ g/mol = 1.5 \ g$.Total mass of the mixture $= 0.8 \ g + 1.5 \ g = 2.3 \ g$.

$A$ gaseous mixture contains $CH_4$ and $C_2H_6$ in equimolecular proportion. The weight of $2.24 \ L$ of this mixture at $S.T.P.$ is :- .............. $g$

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