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

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 $=0.0125 \times 10^{-2} \mathrm{~m}$

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