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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.
Solution
(i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
∴ Mass percent of 15 ppm chloroform in water
`=15/10^6xx100`
`≃ 1.5 xx10^-3 %`
(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
`=(1.5xx10^-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
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