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The Van’tHoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is
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a
91.3%
b
87%
c
100%
d
74%
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Correct option is B
n=3i=2.74i= 1+α(n-1)2.74 = 1+α(3-1)=1+2α2α=2.74-1=1.74α=1.742=0.87 Degree of ionization of 0.1M Ba(NO3)2 = 0.87percentage of ionization of 0.1M Ba(NO3)2 = 87%Similar Questions
The values of van't Hoff factors for KCl, NaCl and K2SO4 respectively are
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