A sample of drinking water was found to be severely contaminated with chloroform, `CHCl_(3)`, supposed to be carcinogen. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent 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`