If the mole fraction of the solute in benzene | vedclass

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(A) The formula relating molality $(m)$ and mole fraction $(X_2)$ is: $m = \frac{1000 	imes X_2}{M_1 	imes X_1}$ Where $X_2$ is the mole fraction of

Vedclass · Accepted Answer

$3.2$

Vedclass · Answer

(A) The formula relating molality $(m)$ and mole fraction $(X_2)$ is: $m = \frac{1000 	imes X_2}{M_1 	imes X_1}$ Where $X_2$ is the mole fraction of the solute,$X_1$ is the mole fraction of the solvent (benzene),and $M_1$ is the molar mass of benzene $(C_6H_6 = 78 \ g/mol)$. Given $X_2 = 0.2$,then $X_1 = 1 - 0.2 = 0.8$. Substituting the values: $m = \frac{1000 	imes 0.2}{78 	imes 0.8} = \frac{200}{62.4} \approx 3.205 \ m$. Thus,the correct option is $3.2$.

If the mole fraction of the solute in benzene is $0.2$,then the molality of the solution will be ..............

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