A sample of drinking water was found to be severely contaminated with chloroform, CHCl_(3) supposed to be carcinogenic in nature. The level of contamination was 15 ppm ( by mass). (i) Express this in precent 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 carcinogenic in nature. The level of contamination was 15 ppm ( by mass). (i) Express this in precent by mass. (ii) Determine the molality of chloroform in the water sample.
Answer» Solution :(i) 15 PPM means 15 parts of million, So, Mass percent `= (15xx100)/(10^(6))=1.5xx10^(-3)%` (ii) Molecular mass of `CHCl_(3)` `= 12.01+1.0079+(3xx35.45)` `= 119.367 ~=119` gm/mol `1.5xx10^(-3)%` means, `1.5xx10^(-3)` gm chloroform in 100 gm Molality `= ("Weight of solute " xx 1000)/("Molecular mass of solute" xx "VOLUME of solution (in gm)")` `=(1.5xx10^(-3) xx 1000)/(119xx100)` `= 0.000126=1.26xx10^(-4)m`