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 percent by mass.
2. Determine the molality of chloroform in the water sample.

Answer 1

(i) 1 ppm is equivalent to 1 part out of 1 million (10⁶) parts.

∴ Mass percent of 15 ppm chloroform in water

= `15/10^6 xx 100`

≈ 1.5 × 10⁻³ g

(ii) 100 g of the sample contains 1.5 × 10^–3 g of CHCl₃

⇒ 1000 g of the sample contains 1.5 × 10^–2 g of CHCl₃

∴ Molality of chloroform in water

= `(1.5 xx 10^-2  g)/("Molar mass of CHCl"_3)`

Molar mass of CHCl₃ = 12.00 + 1.00 + 3(35.5)

= 119.5 g mol^–1

∴ Molality of chloroform in water = 0.0125 × 10^–2 m

= 1.25 × 10^–4 m

Answer 2

i. Since the concentration of CHCl₃ in the drinking water sample is 15 ppm (by mass), 10⁶ g of the sample would contain 15 g of CHCl₃.

∴ Mass of pure water = 10⁶ − 15

= 999985 g

Hence, percentage of CHCl₃ by mas = `15/999985 xx 100`

= 1.5 × 10⁻³ g

ii. Molality m is given by w' = `(m xx M' xx w)/1000`

or `m = (w' xx 1000)/(M' xx w)`

= `(15 xx 1000)/(119.5 xx 999985)`

(Mol. mass of CHCl₃ = 119.5) = 1.25 × 10^–4 mol kg^–1