Question

A sample of drinking water was found to be severely contaminated with chloroform supposed to be a cacinogen. the level of contamination was 15ppm. determine the molality of chloroform in water sample

Answer 1

1) 15 ppm means 15 parts in 1 millions (10^6)
and percentage always for 100

hence 15 ppm = (15/10^6 )*100
= 15*10^-4
= 1.5*10^-3
2) molar mass of chloroform = 118.5g/ mol

from percentage by mass 100 gram of sample has 1.5*10^-3 g of chloroform
then 1000 gram of sample has = 1.5*10^-3 g * 1000 gram/ 100 gram
= 1.5*10^-2 g

molality = moles of solute / mass of solvent in Kg
moles of solute = mass of solute / molar mass
= 1.5*10^-2/ 118.5
=1.266*10^-4 moles

molality = 1.266*10^-4 moles/ 1 Kg =1.266*10^-4 m
Thanks & Regards
Rinkoo Gupta
AskIITians faculty

Answer 2

Hey !!

Let the mass of solution be 10⁶ g.

Mass of solute, chloroform = 15 g

(1) % by mass of chloroform = Mass of chloroform / Mass of solution × 100

             = 15 g / 10⁶ g  × 100

             = 15 × 10⁻⁴ %

(2) Mass of solvent, water = 10⁶ g - 15 g ≈ 10⁶ ≈ 10³ kg

 Number of moles of chloroform, CHCl₃ = Mass of chloroform / Molar mass

                = 15 g / 119.5 g mol⁻¹  

                = 0.126 mol

         Molality of solution = Number of moles of chloroform / Mass of water (in kg) = 0.126 mol / 10³ kg

FINAL RESULT = 1.26 × 10⁻⁴  mol kg⁻¹