A solution of a solute in benzene has a molal | vedclass

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(A) Given: Mole fraction of solute $(x_2)$ = $0.2$. Therefore,mole fraction of solvent (benzene,$x_1$) = $1 - 0.2 = 0.8$. Molality $(m)$ is given by t

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$3.2$

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(A) Given: Mole fraction of solute $(x_2)$ = $0.2$. Therefore,mole fraction of solvent (benzene,$x_1$) = $1 - 0.2 = 0.8$. Molality $(m)$ is given by the formula: $m = \frac{x_2 	imes 1000}{x_1 	imes M_1}$,where $M_1$ is the molar mass of the solvent (benzene,$C_6H_6$). Molar mass of benzene $(M_1)$ = $(6 	imes 12) + (6 	imes 1) = 78 \ g/mol$. Substituting the values: $m = \frac{0.2 	imes 1000}{0.8 	imes 78} = \frac{200}{62.4} \approx 3.205 \ m$. Rounding to the nearest option,$X \approx 3.2$.

$A$ solution of a solute in benzene has a molality of $X \ m$. If the mole fraction of the solute is $0.2$,what is the value of $X$?

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