Discussion Forum



1)

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


A) 0.87

B) 0.74

C) 0.91

D) 87

Answer:

Option A

Explanation:

 Given, 

     Molarity =0.1 M

 vant Hoff factor (i)=2.74

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

$Ba(NO_{3})_{2}\rightleftharpoons Ba^{2+}+2NO_{3}^{-}$

 Number of particles dissociated (n)=3

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

                                                              = $\frac{(2.74-1)}{(3-1)}=0.87$