An X molal solution of a compound in benzene | vedclass

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(B) Molality $(X)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.Let the number of moles of solute be $n_2 = X$ and the mass o

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$3.2$

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(B) Molality $(X)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.Let the number of moles of solute be $n_2 = X$ and the mass of solvent (benzene,$C_6H_6$) be $1000 \ g$.The molar mass of benzene is $78 \ g/mol$.Number of moles of solvent $(n_1)$ = $\frac{1000}{78} \approx 12.82 \ mol$.The mole fraction of solute $(x_2)$ is given by $\frac{n_2}{n_1 + n_2} = 0.2$.Substituting the values: $\frac{X}{X + 12.82} = 0.2$.$X = 0.2X + 2.564$.$0.8X = 2.564$.$X = \frac{2.564}{0.8} = 3.205 \approx 3.2$.

An $X$ molal solution of a compound in benzene has a mole fraction of solute equal to $0.2$. The value of $X$ is

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