The molality of an aqueous solution of any so | vedclass

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(C) Given,mole fraction of solute $(x_B)$ = $0.25$. Since it is an aqueous solution,the mole fraction of solvent (water,$x_A$) = $1 - 0.25 = 0.75$. Mo

Vedclass · Accepted Answer

$18.52$

Vedclass · Answer

(C) Given,mole fraction of solute $(x_B)$ = $0.25$. Since it is an aqueous solution,the mole fraction of solvent (water,$x_A$) = $1 - 0.25 = 0.75$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. $m = \frac{n_B}{W_A 	ext{ (in kg)}} = \frac{n_B}{n_A 	imes M_A 	imes 10^{-3}}$,where $M_A$ is the molar mass of water $(18 \ g/mol)$. $m = \frac{x_B}{x_A 	imes M_A 	ext{ (in kg/mol)}} = \frac{0.25}{0.75 	imes 18 	imes 10^{-3}}$. $m = \frac{0.25}{0.0135} \approx 18.52 \ m$.

The molality of an aqueous solution of any solute having a mole fraction of $0.25$ is . . . . . . . (in $m$)

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