Question

If the degree of dissociation of a salt A2B3 be X(beta), then what will be its Van't Hoff factor?

Answer 1

Answer:

4x+1

Explanation:

A2B3,n=5

x=i-1/n-1

x=i-1/4

i=4x+1

Answer 2

Given: The degree of dissociation $A_2B_3$ be X.

To find: We have to find Van't Hoff factor.

Solution:

The vant Hoff 'i' factor of the salt $A_2B_3$ can be determined by the following steps-

The salt $A_2B_3$ dissociates as-

$A_2B_3\rightarrow 2A^{3+}+3B^{2-}$

The degree of dissociation is X.

So, we can write the dissociation of $A_2B_3$ is (1-X).

The formation of the ions is 2X and 3X.

So, we can write-

$i = (1 - x) + 2x + 3x \\ i = 4x + 1$

The value of 'i' is 4X+1.

So, Vant Hoff factor is 4X+1.