Q- 9 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.
Answer
part (i)
In this problem level is 15 ppm by mass
So ppm is the
part per millions
15 ppm = mass of
solute *10^6/mass of solvent ………..(1)
Mass % CHCl3 = mass of solute*10^2 / mass of solvent ...(2)
Divide equation second by first we get
Mass % of
CHCl3 =
15* 10^-4 %
part 2 (ii)
Formula
Molality = number of mole of solute / mass of
solvent in Kg...(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
Number of moles of solute = mass of solute / molar
mass …(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*10^4/119.5
=1.25*10^-5
Now plug the value of number of moles and mass of solvent
in equation (4)
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
1.25*10^-4 m
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