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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).
- Express this in percent by mass.
- Determine the molality of chloroform in the water sample.
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Solution 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
Solution 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
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