SELECT * FROM question_mgmt as q WHERE id=1565 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Solutions a sample of drinking water was found to be severel

Question:

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 mass

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

Answer:

1) 15 ppm means : 15 parts per million(106) of the solutions

So, Percent by mass = (mass of chloroform / total mass) x 100

= (15 / 106) x 100

= 1.5 x 10-3 %

 

2) Molality Mass of chloroform = 15 g

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

= 119.5 g mol - 1

Moles of chloroform = 15 / 119.5 = 0.1255 mol

Mass of water = 106

Therefore molality = (moles of chloroform / mass of water ) x 1000

= (0.1255 / 106) x 1000

= 1.255 x 10-4 m


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