A sample of drinking water was found to be severely contaminated with Chloroform, CHCl3 . The level of contamination was 15 ppm(by mass).(i) Express this in percent by mass. (ii) Determine the molality of chloroform in water.

1).Converting 15ppm in percent by mass-
15ppm means 15 parts in 1 million(10⁶)
and % means 100
so, 15 ppm = 15 * 10^2/10^{6
= 1.5 * 10-3
2).For Molality-
Molality = Moles of solute/Mass of solvent in Kg
now,
100gm of sample contains 1.5 * 10-3 of CHCl3
therefore, 1000gm will contains = 1.5 * 10-3 * 1000/100
= 1.5.* 10-2g
Moles of solute is given as = Mass/Molar mass
= 1.5 *10-2/118.5 ( 118.5 Molar mass of CHCl3)
moles = 1.266 * 10-4 moles
Thus molality = 1.266 * 10-4/1 Kg
= 1.266 * 10-4m
Thanks & Regards
Aarti Gupta
askiitians Faculty}

Here, ppm = parts per million = 15. Using %w/w = ppm x 10^-4, %w/w = 15/10000 = 0.0015 % Now take 1 million gm soln. Wt. solute = 15 g. This means moles of solute = 15/120(approx) = 0.125 mol. Now wt. of solvent is approx. 10^6 gm. Molality = moles of solute x 1000 / wt solvent in gm = 0.125/1000 =0.000125 m or 1.25 x 10^-3 m

15 ppm = 15/ 10^6(As 15 ppm means 15 parts in 1 million)I) percentage means 100Mass % = 15*100/10^6 =1.5*10^-32) 100g contains 1.5*10^-3Then 1000g will contain 1.5*10^-2Moles = 1.5*10^-2/119.5 = 1.2*10^-4Molality = 1.2*10^-4/ 1kg =1.2*10^-4 molar

Hello students,
The solution is as follows
1).Converting 15 ppm in percent by mass-
15 ppm means 15 parts in 1 million(10^6) and % means 100
so, 15 ppm = 15 * 10^2/10^6
= 1.5 * 10^-3
2).For Molality-
Molality = Moles of solute/Mass of solvent in Kg
now, 100gm of sample contains 1.5 * 10-3 of CHCl3
therefore, 1000gm will contains = 1.5 * 10-3 * 1000/100 = 1.5.* 10-2g
Moles of solute = Mass / Molar mass
= 1.5 * 10^-2 / 118.5 ( 118.5 Molar mass of CHCl3)
moles = 1.266 * 10^-4 moles
Thus molality = 1.266 * 10^-4/1 Kg
= 1.266 * 10^-4m
Ihope the avove solution will solve all your doubts.
Thanks
All the best for your exams.