Explanation
1 ppm=1 mg/L=10−3g/L
Water sample contains 1 ppm urea concentration.
∴ 10−3 g of urea in 1 L
x g urea in 0.05×10−3 L
x=0.05×10−6 g
60 g urea =6.023×1023 molecules
0.05×10−6 g=n molecules
n=6.023×1023×0.05×10−660
=0.005×1017
=5×1014 molecules
We know that
Ka=KwKb=10−141.7×10−9=5.88×10−6
C5H5N+H2O⇌C5H5NH+C5H5NH++OH−
Initial conc: 0.1 0 0 0
Final Conc: 0.1−x x x
Ka=x20.1−x [Since x<<0.1 it can be neglected]
5.88×0.1×10−6=x2
x=7.6×10−4
%Pyridine=7.6×10−40.1×100=0.77%
∵(dissociated ionsconcentration of solution×100)
Hence, the correct option is A
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