plzz solve question no. 103
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C6H5OH → 1/2(C6H5OH)2
here you can see that two C6H5OH molecule associate to form one molecule .
we also write above equation ,
C6H5OH + C6H5OH →(C6H5OH)2
number of particles in the solution from 1 mole solute = 1/2
so,
van't hoff factor ( i ) = 1- 1/2ß
where ß is degree of association .
given,
ß = 0.4
so,
i = 1 - 1/2(0.4) = 1- 0.2 = 0.8
option (2) is correct .
here you can see that two C6H5OH molecule associate to form one molecule .
we also write above equation ,
C6H5OH + C6H5OH →(C6H5OH)2
number of particles in the solution from 1 mole solute = 1/2
so,
van't hoff factor ( i ) = 1- 1/2ß
where ß is degree of association .
given,
ß = 0.4
so,
i = 1 - 1/2(0.4) = 1- 0.2 = 0.8
option (2) is correct .
abhi178:
now see
ß=0.4??
i=1+ß(N-1)
ß = 40% by fractional 40/100 =0.4
ok got it
understood..... thnks
your formula is dissociation not association
yes.....u are right.... I did this mistake
:-)
:--)
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