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).

By BYJU'S Exam Prep

Updated on: September 25th, 2023

(i) Express this in percentage by mass.

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

Solution:

(i) In percentage by mass it is 0.0015%

(ii) The molality of chloroform in the water sample is 1.266 x 10-4 m

Part (i): determining the proportion using mass:

The amount of solute present in a 10 solution is known as the ppm by mass.

Given that concentration = 15ppm

Now, 106 g of solution contains = 15 g of CHCl3

Hence; 100g of solution contain = 15/106 x 100

= 15 x 10-4

= 0.0015%

Table of content

Part (ii): Determining the chloroform’s molality:

Molar mass of chloroform = 119.5 g/mol

Mass of solvent = 1000000 – 15 = 999985 g

Molality = moles of solute/ moles of solvent in g x 1000

Substituting the values we get:

m = 15/ (119.5 x 999985) x 1000

m = 1.266 x 10-4 m

used in dentistry during root canal operations as an anesthetic.
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In the past, chloroform was used as a laundry solvent and as an extraction dissolvable for fats, greases, oils, and other goods.
Used as a food contact substance and an indirect food additive in adhesive components of food packaging materials.

Summary:

(i) In percentage by mass it is 0.0015%.

(ii) The molality of chloroform in the water sample is 1.266 x 10-4 m.