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.
(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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 17: A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to....
Comments
Helpul.
Thnx
Overall,the method of solving problem is very easy and worth the explanation....
All fine but the Molarity symbol
All fine but the symbol ð
Good but learn molality symbol
I didn't understand
Solved nicely but M is not a symbol of the molality it should be m
Really saral . Means very easy and helpful
That's the answer I was looking for all along Thank you Saralstudy