Wednesday, 1 May 2013

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.


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
(i)
In this problem level is 15 ppm by mass
So   ppm is the part per millions
15 ppm =  mass of solute ×106/mass of solvent                      ...(1)
Mass % = mass of solute × 102 / mass of solvent                     ... (2)

Divide equation second by first we get
Mass %  of CHCl3  =  15 × 10-4 %
(ii)
                                             ... (3)

Here we have already find the mass % is 15 × 10-4 
When even mass % is given take total mass of solution = 100 gram
And mass of solute (CHCl3)  will  =  15 × 10-4  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

Now we have to find the number of moles of solute

                            … (4)
So we have to calculate the molar mass of solute (CHCl3)
Molar mass of CHCl3               =12 +1+106.5
                                                =119.5
Plug the value in equation (4)
No of moles of chloroform        = 15×104/119.5
                                                =1.25×10-5
Now plug the value of number of moles and mass of solvent in equation (4)
 We get
Here mass of solvent = 100 g
Convert it in Kg we divide by 100 we get
Mass of solvent = 0.1Kg 

Molality of            = 1.25×10-5/0.1
                             = 1.25×10-4 m

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