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(C) Since the mixture is equimolecular,the number of moles of $CH_4$ and $C_2H_6$ are equal.Total volume of the mixture $= 2.24 \ L$.At $NTP$,$22.4 \

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$2.3$

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(C) Since the mixture is equimolecular,the number of moles of $CH_4$ and $C_2H_6$ are equal.Total volume of the mixture $= 2.24 \ L$.At $NTP$,$22.4 \ L$ of any gas contains $1 \ mole$.Therefore,total moles in the mixture $= \frac{2.24 \ L}{22.4 \ L/mol} = 0.1 \ mol$.Since they are in equimolecular proportion,moles of $CH_4 = 0.05 \ mol$ and moles of $C_2H_6 = 0.05 \ mol$.Molar mass of $CH_4 = 12 + (4 	imes 1) = 16 \ g/mol$.Molar mass of $C_2H_6 = (2 	imes 12) + (6 	imes 1) = 30 \ g/mol$.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 weight 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 $NTP$ is ............ $g$.

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