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Question 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.

Answer

(i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.

∴ Mass percent of 15 ppm chloroform in water

 

 

(ii) molality (M) = no of moles of solute/mass of solvent in g *1000

 

Therefore mass of chloroform= 12 + 1+3(35.5) = 119.5 g/mol

 

100 g of the sample contains 1.5 × 10–3 g of CHCl3.

 

⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.

 

 m = 1.5 x 10-3/119.5 * 1000 = 1.25x 10-4 m

 

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