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A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogen. The level of contamination was 15 ppm (by mass)(i) Express this in percent by mass(ii) Determine the molarity of chloroform in the water sample. |
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Answer» We know that, ppm = {Mass of chloroform}/{Mass of solution} x 106 Given ppm of chloroform = 15 If the mass of solution is 1 kg (1000 gm) the mass of chloroform = {15 x 1000}/{106} = 0.015 g It means 1000 g of solution contains 0.015 g of chloroform Thus the % of the solution = {0.015 x 100}/{1000} = 0.0015% Because the solution is very dilute and the density of water is 1.0 g mL-1, therefore the mass of the solution will be nearly same to volume of the solution, thus Molarity = {Mass of chloroform}/{Molar mass of chloroform} x {1000}/{Volume of solution} = {0.015}/{119.5} x {1000}/{1000} = 12.6 x 10-5 m = 1.25 x 10-4 m |
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