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A sample of drinking water was found to be severely contaminated with chloroform, `CHCl_(3)`, supposed to be carcinogen. 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. |
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Answer» Correct Answer - `a. 15xx10^(-4)g, b. 1.25xx10^(-25)m` a. `pp m=(W_(2))/("Weight of solution")xx10^(6)` `(15)/(10^(6))=(W_(2))/(Weight of solution)` `%` By mass `=(15)/(10^(6))xx100=15xx10^(-4)g` b. `Mw` of `CHMCl_(30=119.5g` `W_(2)=15xx10^(-4)g` Weight of solution `=100g` Weight of solvent `=W_(1)=(100-15xx10^(-4))` `=99.998g` `m=(W_(2)xx1000)/(Mw_(2)xxW_(1))=(15xx10^(-4)xx10^(3))/(119.5xx99.998)` `=1.25xx10^(-5)m` |
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