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NCERT Class XII Chemistry
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Question : 9 of 53
Marks: +1, -0
A sample of drinking water was found to be severely contaminated with chloroform,CHCl3\mathrm{CHCl}_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.
Solution:  
(i) 15 ppm means 15 parts in million (10 6^{6} ) parts by mass in the solution.
%\% by mass =15106×100=1.5×10−3= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3}
(ii) Mass of solvent =106g= 10^{6} \mathrm{g} (Mass of solute is negligible)
Molar mass of CHCl3=12+1+3×35.5=119.5 g mol−1\mathrm{CHCl}_3 = 12 + 1 + 3 \times 35.5 = 119.5\ \mathrm{g}\ \mathrm{mol}^{-1}
No. of moles of CHCl3=Mass in gMolar mass\mathrm{CHCl}_3 = \frac{ \text{Mass in } \mathrm{g} }{ \text{Molar mass} } =15119.5= \frac{15}{119.5}
∴  \therefore\; Molality =15/119.5106×1000=1.25×10−4 m= \frac{15/119.5}{10^{6}} \times 1000 = 1.25 \times 10^{-4} \ \mathrm{m}
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