Question

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 15ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.

Answer 1

(i) 15 ppm (by mass) means 15 parts per million $(10 ^6 )$ of the solution.
Therefore, percent by mass $=(\frac{15}{10^6})\times100\%$
$= 1.5 \times 10^{ - 3} \%$
(ii) Molar mass of chloroform $(CHCl_3) = 1\times12 + 1\times1 + 3\times35.5$
$= 119.5 \,g mol^{ - 1}$
Now, according to the question,
15 g of chloroform is present in $10^6 g$ of the solution.
i.e., 15 g of chloroform is present in $(10^6 - 15) ≈ 10^6 g$ of water.
Molality of the solution $=\frac{(\frac{15}{119.6} mol) }{(10^6\times10^{-3})}$
$=1.26 × 10^ {- 4} m$