The Van’tHoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is

The degree of dissociation, α=i-1n-1Ba(NO3)2⇌Ba+2+2NO3- ; n=3 α=2.74-13-1=1.742=0.87 Percent of dissociation=87%

The Van’tHoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is

The degree of dissociation, α=i-1n-1Ba(NO3)2⇌Ba+2+2NO3- ; n=3 α=2.74-13-1=1.742=0.87 Percent of dissociation=87%

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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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The degree of dissociation, $α = \frac{i - 1}{n - 1}$

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{+ 2} + 2 {NO}_{3}^{-}; n = 3 α = \frac{2.74 - 1}{3 - 1} = \frac{1.74}{2} = 0.87 Percent of dissociation = 87 %$

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The Van’tHoff factor for 0.1 M Ba(NO₃)₂ solution is 2.74. The degree of dissociation is