Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Mass percentage of $\mathrm{C}_{6} \mathrm{H}_{6}=\frac{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}}{\text { Total mass of the solution }} \times 100 \%$
$=\frac{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}}{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}+\text { Mass of } \mathrm{CCl}_{4}} \times 100 \%$
$=\frac{22}{22+122} \times 100 \%$
$=15.28 \%$
Mass percentage of $\mathrm{CCl}_{4}=\frac{\text { Mass of } \mathrm{CCl}_{4}}{\text { Total mass of the solution }} \times 100 \%$
$=\frac{\text { Mass of } \mathrm{CCl}_{4}}{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}+\text { Mass of } \mathrm{CCl}_{4}} \times 100 \%$
$=\frac{122}{22+122} \times 100 \%$
$=84.72 \%$
Alternatively,
Mass percentage of CCl4 = (100 − 15.28)%
= 84.72%