A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) 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.
(i) $15 \mathrm{ppm}$ (by mass) means 15 parts per million $\left(10^{6}\right)$ of the solution.
Therefore, percent by mass $=\frac{15}{10^{6}} \times 100 \%$
= 1.5 × 10−3 %
(ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5 g mol−1
Now, according to the question,
15 g of chloroform is present in 106 g of the solution.
i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water.
$\therefore$ Molality of the solution $=\frac{\frac{15}{119.5} \mathrm{~mol}}{10^{6} \times 10^{-3} \mathrm{~kg}}$
= 1.26 × 10−4 m