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 percent by mass (ii) Determine the molarity 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 carcinogen. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass (ii) Determine the molarity of chloroform in the water sample.
Answer» SOLUTION :(i) 15 ppm means 15 parts in million `(10^(6))` parts `therefore""%" by mass "=(15)/(10^(6))xx100=15xx10^(-4)=1.5xx10^(-3)%` (ii) Molar mass of chloroform `(CHCl_(3))=12+1+3xx35.5=118.5g mol^(-1)` `"100 g of the SAMPLE CONTAIN chloroform"=1.5xx10^(-3)g` `therefore"1000 g (1 kg) of the sample will contain chloroform "=1.5xx10^(-2)g=(1.5xx10^(-2))/(118.65)" mole"` `=1.266xx10^(-4)" mole"` `therefore"Molality "=1.266xx10^(-4)m`.