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A sample of drinking water was found to be severely contaminated with chloroform, \(\mathrm{CHCl}_3\), 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) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
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Verified Answer
(i) \(15 \mathrm{ppm}\) means 15 parts in million \(\left(10^6\right)\) parts.
\(\%\) by mass \(\frac{15}{10^6} \times 100=15 \times 10^{-4}=1.5 \times 10^{-3} \%\)
(ii) Molar mass of chloroform
\(\left(\mathrm{CHCl}_3\right)=12+1+3 \times 35.5=119.5 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(100 \mathrm{~g}\) of the sample contain chloroform \(=1.5 \times 10^{-3} \mathrm{~g}\)
\(\therefore 1000 \mathrm{~g}(1 \mathrm{~kg})\) of the sample will contain chloroform \(=1.5 \times 10^{-2} \mathrm{~g}\)
Now molality \(=\frac{1.5 \times 10^{-2}}{119.5}=1.255 \times 10^{-4} \mathrm{~m}\)
\(\therefore\) Molality \(=1.255 \times 10^{-4} \mathrm{~m}\).
\(\%\) by mass \(\frac{15}{10^6} \times 100=15 \times 10^{-4}=1.5 \times 10^{-3} \%\)
(ii) Molar mass of chloroform
\(\left(\mathrm{CHCl}_3\right)=12+1+3 \times 35.5=119.5 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(100 \mathrm{~g}\) of the sample contain chloroform \(=1.5 \times 10^{-3} \mathrm{~g}\)
\(\therefore 1000 \mathrm{~g}(1 \mathrm{~kg})\) of the sample will contain chloroform \(=1.5 \times 10^{-2} \mathrm{~g}\)
Now molality \(=\frac{1.5 \times 10^{-2}}{119.5}=1.255 \times 10^{-4} \mathrm{~m}\)
\(\therefore\) Molality \(=1.255 \times 10^{-4} \mathrm{~m}\).
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