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(C) The mixture contains $CH_4$ and $C_2H_6$ in equimolecular proportion,meaning the mole fraction of each is $0.5$.The average molar mass of the mixt

Vedclass · Accepted Answer

$2.3$

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(C) The mixture contains $CH_4$ and $C_2H_6$ in equimolecular proportion,meaning the mole fraction of each is $0.5$.The average molar mass of the mixture is calculated as: $M_{avg} = (0.5 	imes M_{CH_4}) + (0.5 	imes M_{C_2H_6}) = (0.5 	imes 16) + (0.5 	imes 30) = 8 + 15 = 23 \ g/mol$.At $NTP$,$22.4 \ L$ of any gas corresponds to $1 \ mole$.Therefore,$2.24 \ L$ of the mixture corresponds to $0.1 \ mole$.The weight of the mixture is: $Weight = 	ext{moles} 	imes M_{avg} = 0.1 	imes 23 = 2.3 \ g$.

$A$ 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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