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.

A sample of drinking water was found to be severely contaminated with

1.	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.
Answer» SOLUTION :(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`.