Calculate the mole fraction of benzene in solution containing $30 %$ by mass in carbon tetrachloride.

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Solution

$C l_{4} (n) = \frac{70}{154} = 0.455$

$n C_{6} H_{6} = \frac{n C_{6} H_{6}}{n C_{6} H_{6} + n C C l_{4}}$

$= \frac{0.385}{0.385 + 0.455}$

$= \frac{0.385}{0.84}$

= 0.458

$n C C l_{4} = 1 - 0.458$

= 0.542

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