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» <html><body><p> <br/> <br/> </p>Solution :(i) 15 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a> means 15 parts of million, <br/> So, Mass percent `= (15xx100)/(10^(6))=1.5xx10^(-3)%` <br/> (ii) Molecular mass of `CHCl_(3)` <br/> `= 12.01+1.0079+(3xx35.45)` <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/119-268764" style="font-weight:bold;" target="_blank" title="Click to know more about 119">119</a>.367 ~=119` gm/mol <br/> `1.5xx10^(-3)%` means, `1.5xx10^(-3)` gm chloroform in 100 gm <br/> Molality `= ("Weight of solute " xx 1000)/("Molecular mass of solute" xx "<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of solution (in gm)")` <br/> `=(1.5xx10^(-3) xx 1000)/(119xx100)` <br/> `= 0.000126=1.26xx10^(-4)m`</body></html>