$A$ sample of drinking water was found to be severely contaminated with chloroform $(CHCl_{3})$,supposed to be a carcinogen. The level of contamination was $15$ $ppm$ (by mass).
$(i)$ Express this in percent by mass.
$(ii)$ Determine the molality of chloroform in the water sample.

(N/A) $15$ $ppm$ means $15$ parts in $10^{6}$ parts by mass in the solution.
$(i)$ Percentage by mass $= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3} \%$.
$(ii)$ As $15 \, g$ of chloroform is present in $10^{6} \, g$ of the solution,the mass of the solvent is approximately $10^{6} \, g$ (since $15 \, g$ is negligible compared to $10^{6} \, g$).
Molar mass of $CHCl_{3} = 12 + 1 + 3 \times 35.5 = 119.5 \, g \, mol^{-1}$.
Moles of $CHCl_{3} = \frac{15 \, g}{119.5 \, g \, mol^{-1}} \approx 0.1255 \, mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1255 \, mol}{10^{6} \, g / 1000} = \frac{0.1255}{1000} \, mol \, kg^{-1} = 1.255 \times 10^{-4} \, m$.