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 \(=\frac{(\frac{15}{119.6} mol) }{(10^6\times10^{-3})}\)
\(=1.26 × 10^ {- 4} m\)