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  4. If the van’t Hoff-factor for 0.1 M Ba(NO3)2 solution is 2.74, the degree of dissociation is

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Q. If the van’t Hoff-factor for $0.1\, M\, Ba(NO_3)_2$ solution is $2.74$, the degree of dissociation is

MHT CETMHT CET 2019

Solution:

Given,

Molarity $=01 M$ vant Hoff factor $(i)=2.74$

Since, $i>1$, it means solute is undergoing dissociation.

$Ba \left( NO _{3}\right)_{2} \rightleftharpoons Ba ^{2+}+2 NO _{3}^{-}$

Number of particles dissociated $(n)=3$

Now, $\alpha$ (degree of dissociation) $=\frac{(i-1)}{(n-1)}$

$=\frac{(2.74-1)}{(3-1)}$

$=0.87$