# A sample of drinking water was foundto be severely contaminated with chloroform,

Question.

A sample of drinking water was found to be severely contaminated with chloroform, $\mathrm{CHCl}_{3}$,

supposed to be 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.

Solution:

(i) $1 \mathrm{ppm}$ is equivalent to 1 part out of 1 million $\left(10^{6}\right)$ parts.

Mass percent of 15 ppm chloroform in water

$=\frac{15}{10^{6}} \times 100$

$\simeq 1.5 \times 10^{-3} \%$

(ii) $100 \mathrm{~g}$ of the sample contains $1.5 \times 10^{-3} \mathrm{~g}$ of $\mathrm{CHCl}_{3}$.

$\Rightarrow 1000 \mathrm{~g}$ of the sample contains $1.5 \times 10^{-2} \mathrm{~g}$ of $\mathrm{CHCl}_{3}$.

$\therefore$ Molality of chloroform in water

$=\frac{1.5 \times 10^{-2} \mathrm{~g}}{\text { Molar mass of } \mathrm{CHCl}_{3}}$

Molar mass of $\mathrm{CHCl}_{3}=12.00+1.00+3(35.5)$

$=119.5 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Molality of chloroform in water $=0.0125 \times 10^{-2} \mathrm{~m}$

$=1.25 \times 10^{-4} \mathrm{~m}$

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