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Q.

Osmotic pressure of 0.1 M monoprotic acid solution at 300K is 4.926 atm .  Percentage of ionization of the acid   is

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a

75%

b

50%

c

60%

d

40%

answer is B.

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Detailed Solution

π=iCST 4.926=i×0.1×0.0821×300 i=4.9262.463 i = 2 Na2SO42Na++SO42 n =3 i=1α+ i=1+2α 2=1+2α α=0.5

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