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Exercises · 1.9

Q.A sample of drinking water was found to be severely contaminated with chloroform (CHCl3CHCl_3) 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.
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Working on a clean 1 kg1\ \text{kg} basis, 1515 ppm chloroform gives 1.5×10−3 %1.5 \times 10^{-3}\ \% by mass and a molality of 1.25×10−4 m1.25 \times 10^{-4}\ m.

Step 1 — Read what "ppm by mass" means.

Parts per million by mass tells us the mass of solute present in every 10610^{6} mass units of solution. So 1515 ppm means

15 ppm=15 g of CHCl3106 g of solution15\ \text{ppm} = \frac{15\ \text{g of }\mathrm{CHCl_3}}{10^{6}\ \text{g of solution}}

Step 2 — (i) Convert to percent by mass.

Percent by mass is parts per hundred, so multiply the mass fraction by 100100:

% by mass=15106×100=1.5×10−3 %\%\ \text{by mass} = \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3}\ \%

Step 3 — Choose a convenient sample size for molality.

Molality needs moles of solute per kilogram of solvent. Take 1 kg=1000 g1\ \text{kg} = 1000\ \text{g} of the solution. Scaling the ppm ratio down to this 1000 g1000\ \text{g} sample:

mass of CHCl3=15106×1000 g=0.015 g\text{mass of }\mathrm{CHCl_3} = \frac{15}{10^{6}} \times 1000\ \text{g} = 0.015\ \text{g}

Since the chloroform is only a trace contaminant, the mass of water (the solvent) is essentially the whole sample:

mass of water≈1000 g=1 kg\text{mass of water} \approx 1000\ \text{g} = 1\ \text{kg}

Step 4 — Moles of chloroform.

The molar mass of CHCl3\mathrm{CHCl_3} is …

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