Question

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

पर्याय

• 0.87

• 0.74

• 0.91

• 87

Answer 1

If the van't Hoff-factor for 0.1 M Ba(NO₃)₂ solution is 2.74, the degree of dissociation is 0.87.

Explanation:

Given,

Molarity = 0.1 M

vant Hoff factor (i) = 2.74

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

\[\ce{Ba(NO3)2 <=> Ba^{2+} + 2NO3}\]

Number of particles dissociated (n) = 3

Now, (degree of dissociation) = `("i" - 1)/("n" - 1)`

`= (2.74 - 1)/(3 - 1)`

= 0.87