The mole fraction of a solvent in an aqueous | vedclass

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(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent}

Vedclass · Accepted Answer

$13.88$

Vedclass · Answer

(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent} = 18 \ g \ mol^{-1}$.Mole fraction of solute $(X_{solute})$ = $1 - X_{solvent} = 1 - 0.8 = 0.2$.Let the total number of moles be $n_{total} = 1$. Then $n_{solute} = 0.2 \ mol$ and $n_{solvent} = 0.8 \ mol$.Mass of solvent = $n_{solvent} 	imes M_{solvent} = 0.8 \ mol 	imes 18 \ g \ mol^{-1} = 14.4 \ g = 0.0144 \ kg$.Molality $(m)$ = $\frac{n_{solute}}{	ext{Mass of solvent in } kg} = \frac{0.2 \ mol}{0.0144 \ kg} = 13.88 \ mol \ kg^{-1}$.

The mole fraction of a solvent in an aqueous solution of a solute is $0.8$. The molality (in $mol \ kg^{-1}$) of the aqueous solution is:

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