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

Explanation

(i) 15 ppm (by mass) means 15 parts per million (106) of the solution.
Therefore, Percent by mass `= (15)/(10^(6))xx100`
`= 1.5xx10^(-3)%`
(ii) Molar mass of chloroform `(CHCl_(3))`
(ii) Molar mass of chloroform `(CHCl_(3))`
`=1xx12+1xx1+3xx35.5`
`= 119.5 g mol^(-1)`
Now, according to the question,
15 g of chloroform is present in 106 g of the solution. 1.e., 15 g of chloroform is present in (106 - 15) per 106 g of water :
Therefore, Molality of the solution
`= ((15)/(119.5)mol)/(10^(6)xx10^(-3)kg)=1.26xx10^(-4)m`.