Question

A sample of drinking water was found to be severely contaminated with chloroform $CH{Cl}_3$, supposed to be carcinogenic in nature. 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 1

15 ppm it means that in ${10}^6$and % for 100

hence 15 ppm=$\frac{15}{{10}^6}\times 100$

molar mass of chloroform=118.5 g/mol

form % by mass 100 g of a sample has $1.5\times {10}^{-3}$ of choloroform

then 1000 g of a sample has $\frac{1.5\times {10}^{-3}\times 1000}{100gram}$

molality=$\frac{molesofsolute}{massofsolventin1kg}$

$\frac{massofsolute}{molarmass}$

$\frac{1.5\times {10}^{-2}}{118.5}$

=$1.266\times {10}^{-4}$ moles