Question

A sample of drinking water was found to be severely contaminated with chloroform $\left(CHC{l}_3\right)$ 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

$15$ ppm corresponds to $15$ g chloroform in $1000,000$ g of solution.

(i) Percent by mass $=\frac{\text{Mass of chloroform}}{\text{Mass of solution}}\times 100$

Percent by mass $=\frac{15}{1000,000}\times 100=1.5\times {10}^{-3}\%$

(ii) Molality $=\frac{\text{Mass of chloroform}}{\text{Molar mass of chloroform}\times (\text{Mass of solution}-\text{mass of chloroform})}\times 1000$ (all masses in g)

Molality $=\frac{15}{119.5\times (1000,000-15)}\times 1000=1.255\times {10}^{-4}$ m.