Question

Question 1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.

Class XI Some Basic Concepts of Chemistry Page 23

Answer 1

(i)
concept : X ppm means in solution , X parts in the one million ( 10^6) parts .

here, 15 ppm
therefore , % by mass = 15 × 100/10^6 = 1.5 × 10^-3 %

(ii)
we know,
Malority = % by mass × density of solⁿ × 10/molar mass of solute .
% by mass = 1.5 × 10^-3
density of solⁿ = 1 g/cm³ [ because solⁿ is water ]
molar mass of solute (CHCl3 ) = 119 g/mol

Molarity = 1.5 × 10^-3 × 1 × 10/119
= 1.26 × 10^-4 M

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