Question

A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

1. Express this in per cent by mass.
2. Determine the molality of chloroform in the water sample.

Answer 1

Given

Level of contamination \( = 15 \,\text{ppm (by mass)} \)
Solute: chloroform, \( \text{CHCl}_3 \)

1. Convert 15 ppm to Percent by Mass

By definition, 15 ppm (by mass) means: \[ 15 \text{ parts of solute in } 10^6 \text{ parts of solution (by mass)} \] Percent by mass is parts per 100: \[ \%\text{ by mass} = \frac{15}{10^6} \times 100 = 1.5 \times 10^{-3}\% \]

Percent by mass of chloroform \( = 1.5 \times 10^{-3}\% \)

2. Molality of Chloroform in Water

(a) Take a Reference Sample

15 ppm by mass \( \Rightarrow \) 15 g \(\text{CHCl}_3\) in \(10^6\) g solution (water + chloroform).
Since contamination is very small: \[ \text{Mass of solvent (water)} \approx 10^6 \text{ g} = 1000 \text{ kg} \]

(b) Moles of Chloroform

Molar mass of \(\text{CHCl}_3\): \[ M(\text{CHCl}_3) = 12 + 1 + 3 \times 35.5 = 119.5 \,\text{g mol}^{-1} \] \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{15}{119.5} \approx 0.125 \,\text{mol} \]

(c) Molality Calculation

Molality \( m = \dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}} \) \[ m = \frac{0.125}{1000} = 1.25 \times 10^{-4} \,\text{mol kg}^{-1} \]

Molality of chloroform in water \( = 1.25 \times 10^{-4} \,\text{m} \).