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 per cent by mass.

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

Asked by Abhisek | 1 year ago |  76

Solution :-

(a) 1 ppm = 1 part out of 1 million parts.

Mass percent of 15 ppm chloroform in H2O

$$\dfrac{15}{10^{6}} \times 100$$

$$\approx1.5 ×10^{-3} %$$

(b) 100 grams of the sample is having $$1.5 ×10^{-3}\;g \;of \;CHCl_{3}$$

1000 grams of the sample is having $$1.5 ×10^{-2}g\; of\; CHCl_{3}$$

$$Molality \;of CHCl_{3}$$ in water

= $$1.5 ×10^{-2}g\; of\; CHCl_{3}$$ 

Molar mass $$CHCl_{3}$$

= 12 + 1 + 3 (35.5)

= 119.5 grams$$mol^{-1}$$

Therefore, molality of $$CHCl_{3}$$

I water

= 1.25 × $$10^{-4}$$ m

Answered by Pragya Singh | 1 year ago

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