Mole fraction of benzene in a $30 \%(w / w)$ solution in carbon tetrachloride is

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Professor PC Thomas Classes · Answer

$n_{1}=\frac{w_{1}}{M_{1}}=\frac{30}{78}=0.38 n_{2}=\frac{w_{2}}{M_{2}}=\frac{70}{154}=0.45 \chi_{1}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{0.38}{(0.38+0.45)}=0.46$