A sample of drinking water was found to e severely contaminated with chloroform `(CHCl_(3))` supposed to e 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

`~15xx10^(-4) g ,1.25 xx10^(-4) m`

Explanation

(i) 1 PPm is equivalent to part out of million `(10^(6))` parts
` therefore ` mass percent of 15 ppm chloroform in water
`=(15)/(10^(6))xx100`
`~- 1.5 xx10^(-3)%`
(ii) 100g of the sammple containts `1.5 xx10^(-3) g of CHCl_(3)`.
`implies` 1000 g of the sample contains `1.5xx10^(-2) g ofCHCl_(3)`
`therefore ` Molality of chloofromn in water
`=(1.5 xx10^(-2) g)/("Molar mass of CHCl_(3))`
Molar mass of `CHCl_(3) =12.00 +1.00 + 3(35.5)`
`=119.5 g mol ^(-1)`
` therefore` Molality of chlorofrom in water `=0.0125xx10^(-2)` m
`=1.25xx10^(-4) m`