$A$ sample of drinking water was found to be severely contaminated with chloroform,$CHCl_{3}$,supposed to be carcinogenic in nature. 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) $(i)$ $1 \ ppm$ is equivalent to $1$ part out of $10^{6}$ parts.
Mass percent of $15 \ ppm$ chloroform in water $= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3} \%$.
$(ii)$ $10^{6} \ g$ of the sample contains $15 \ g$ of $CHCl_{3}$.
Mass of water $\simeq 10^{6} \ g$ (since the amount of solute is negligible).
Molar mass of $CHCl_{3} = 12.01 + 1.008 + 3(35.45) = 119.37 \ g \ mol^{-1}$.
Moles of $CHCl_{3} = \frac{15 \ g}{119.37 \ g \ mol^{-1}} \simeq 0.1256 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.1256 \ mol}{10^{6} \ g \times 10^{-3} \ kg/g} = 1.256 \times 10^{-4} \ m$.