In a solution of 7.8 g benzene (C_6H_6) and | vedclass

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(A) Step $1$: Calculate the number of moles of benzene $(C_6H_6)$.Molar mass of $C_6H_6 = (6 	imes 12) + (6 	imes 1) = 78 \ g/mol$.Moles of $C_6H_6

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$1/6$

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(A) Step $1$: Calculate the number of moles of benzene $(C_6H_6)$.Molar mass of $C_6H_6 = (6 	imes 12) + (6 	imes 1) = 78 \ g/mol$.Moles of $C_6H_6 = \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$.Step $2$: Calculate the number of moles of toluene $(C_6H_5CH_3)$.Molar mass of $C_6H_5CH_3 = (7 	imes 12) + (8 	imes 1) = 92 \ g/mol$.Moles of $C_6H_5CH_3 = \frac{46.0 \ g}{92 \ g/mol} = 0.5 \ mol$.Step $3$: Calculate the mole fraction of benzene $(X_{benzene})$.$X_{benzene} = \frac{	ext{moles of benzene}}{	ext{moles of benzene} + 	ext{moles of toluene}} = \frac{0.1}{0.1 + 0.5} = \frac{0.1}{0.6} = \frac{1}{6}$.

In a solution of $7.8 \ g$ benzene $(C_6H_6)$ and $46.0 \ g$ toluene $(C_6H_5CH_3)$,the mole fraction of benzene in this solution is:

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