A solution contains 7.8 g of benzene (C_6H_6 | vedclass

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(A) Step $1$: Calculate the molar mass of benzene $(C_6H_6)$ and toluene $(C_6H_5CH_3)$.$M(C_6H_6) = (6 	imes 12) + (6 	imes 1) = 78 \ g/mol$.$M(C_6

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$1/6$

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(A) Step $1$: Calculate the molar mass of benzene $(C_6H_6)$ and toluene $(C_6H_5CH_3)$.$M(C_6H_6) = (6 	imes 12) + (6 	imes 1) = 78 \ g/mol$.$M(C_6H_5CH_3) = (7 	imes 12) + (8 	imes 1) = 92 \ g/mol$.Step $2$: Calculate the number of moles of each component.$n(C_6H_6) = \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$.$n(C_6H_5CH_3) = \frac{46.0 \ g}{92 \ g/mol} = 0.5 \ mol$.Step $3$: Calculate the mole fraction of benzene $(x_{benzene})$.$x_{benzene} = \frac{n(C_6H_6)}{n(C_6H_6) + n(C_6H_5CH_3)} = \frac{0.1}{0.1 + 0.5} = \frac{0.1}{0.6} = \frac{1}{6}$.

$A$ solution contains $7.8 \ g$ of benzene $(C_6H_6)$ and $46.0 \ g$ of toluene $(C_6H_5CH_3)$. What is the mole fraction of benzene?

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