134. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. 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.

Answer:

(i) 15 ppm of CHCl₃ in water means that there is 15 g of CHCl₃ in 10⁶ g of water (1 million = 10⁶)
Percentage of CHCl_{3
$$\begin{gathered} = \frac{15}{{10}^6}\times 100 = 0.0015 \% \\ = 1.5 \times {10}^{-4}\% \end{gathered}$$}

(ii)
Here we have already find the mass % is 15 x 10^-4 

 Number of moles of solute = mass of solute / molar mass

number of moles of CHCl₃ =$\frac{15\times {10}^{-4}}{118.5} = 1.266\times {10}^{-5}$

Molality, m of CHCl3 in drinking water sample
            $= \frac{\mathrm{moles} \mathrm{of} {\mathrm{CHCl}}_3}{\mathrm{mass} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{kg}}$