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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by

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प्रश्न

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

  1. express this in percent by mass.
  2. determine the molality of chloroform in the water sample.
संख्यात्मक
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उत्तर

(i) 15 ppm (by mass) means 15 parts per million (106) of the solution.

∴ Percent by mass  = `15/10^6 xx 100`

= 1.5 × 10−3

(ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol−1

Now, according to the question,

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water.

Number of moles of CHCl3 = `"Mass (in g)"/"Molar mass"`

= `15/119.5`

∴ Molality of the solution = `(15/119.5  "mol")/(10^6) xx 1000`

= 1.25 × 10−4 m

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अध्याय 1: Solutions - Exercises [पृष्ठ २८]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 1 Solutions
Exercises | Q 1.9 | पृष्ठ २८

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