A sample of drinking water was found to be severely contaminated with chloroform

(i) $15 \mathrm{ppm}$ (by mass) means 15 parts per million $\left(10^{6}\right)$ of the solution.

Therefore, percent by mass $=\frac{15}{10^{6}} \times 100 \%$

= 1.5 × 10−3 %

(ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol−1

Now, according to the question,

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water.

$\therefore$ Molality of the solution $=\frac{\frac{15}{119.5} \mathrm{~mol}}{10^{6} \times 10^{-3} \mathrm{~kg}}$

= 1.26 × 10−4 m