Calculate the mass percentage of benzene $(C_{6}H_{6})$ and carbon tetrachloride $(CCl_{4})$ if $22 \ g$ of benzene is dissolved in $122 \ g$ of carbon tetrachloride.

Mass percentage of $C_{6}H_{6} = \frac{\text{Mass of } C_{6}H_{6}}{\text{Total mass of the solution}} \times 100 \%$
$= \frac{22 \ g}{22 \ g + 122 \ g} \times 100 \%$
$= \frac{22}{144} \times 100 \% = 15.28 \%$
Mass percentage of $CCl_{4} = \frac{\text{Mass of } CCl_{4}}{\text{Total mass of the solution}} \times 100 \%$
$= \frac{122 \ g}{22 \ g + 122 \ g} \times 100 \%$
$= \frac{122}{144} \times 100 \% = 84.72 \%$
Alternatively,
Mass percentage of $CCl_{4} = (100 - 15.28) \% = 84.72 \%$