$A$ sample of drinking water is severely contaminated with chloroform $(CHCl_3)$. $CHCl_3$ is considered 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) $(i)$ $15 \ ppm$ means $15$ parts per million. Therefore,percent by mass $= \frac{15 \times 100}{10^6} = 1.5 \times 10^{-3} \%$.
$(ii)$ Molar mass of $CHCl_3 = 12.01 + 1.008 + (3 \times 35.45) = 119.37 \ \text{g/mol}$.
$1.5 \times 10^{-3} \%$ means $1.5 \times 10^{-3} \ \text{g}$ of $CHCl_3$ is present in $100 \ \text{g}$ of solution.
Mass of solvent (water) $\approx 100 \ \text{g} = 0.1 \ \text{kg}$.
$\text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.5 \times 10^{-3} \ \text{g} / 119.37 \ \text{g/mol}}{0.1 \ \text{kg}} = 1.256 \times 10^{-4} \ \text{m} \approx 1.26 \times 10^{-4} \ \text{m}$.