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 percentage 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 percentage by mass (ii) determine the molality of chloroform in the water sample.
Answer» <html><body><p></p>Solution :(i) <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> ppm means 15 parts in <a href="https://interviewquestions.tuteehub.com/tag/million-560845" style="font-weight:bold;" target="_blank" title="Click to know more about MILLION">MILLION</a> `(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(6))` parts by mass in the solution <br/> `therefore %` of mass `= (15)/(10^(6) xx 100 = 1.5 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` <br/> (ii) Taking 15 g chloroform in `10^(6)` g of the solution <br/> Mass of the solvent `= 10 ^(6)g` <br/> Molar mass of `CHCl_(3) = 12 + 1 + (3 xx 35.5) = 119.5 g mol^(-1)` <br/> `therefore`Molality `= ((15)/(119.5))/(10^(6)) xx 100 = 1.25 xx 10 ^(-5)m`</body></html>