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Solutions

Question
CBSEENCH12005729

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), 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

Answer:

(i) 15 ppm of CHCl3 in water means that there is 15 g of CHCl3 in 106 g of water (1 million = 106)
Percentage of CHCl3
= 15106×100 = 0.0015 % = 1.5 × 10-4%

(ii)
Here we have already find the mass % is 15 x 10-4 

When even mass % is given take total mass of solution = 100 gram
And mass of solute (CHCl3)  will  =  15 x10-4 g
Mass of solvent = mass of solution – mass of solute
(Here mass of solute is very small as compare to total mass of solution so neglect it and we get)
Mass of solvent = mass of solution
Mass of solvent = 100 g
 

 Number of moles of solute = mass of solute / molar mass

number of moles of CHCl3 =15×10-4118.5 = 1.266×10-5


Molality, m of CHCl3 in drinking water sample
            = moles of CHCl3mass of water in kg

Now plug the value of number of moles and mass of solvent in above eqution
 
Here mass of solvent = 100 g
Convert it in Kg we divide by 100 we get
Mass of solvent = 0.1Kg 
   
molaity = 1.266 x 10-5 / 0.1 = 1.266 x 10-4 mol Kg-1