Question

An aqueous solution of a salt mx2 at certain temperature has a van't hoff factor of 2.The degree of dissociation for this solution of the salt is :

Answer 1

MX2  dissociates into M^2 +2X-  

Given van't hoff factor is equal to 2. Let the initial number of the compound MX2 is 1, the degree of dissociation be y,

MX2 ⇒M^2 + 2X-

1             0        0

1-y        y            2y

van't hoff factor= number of molecules after dissociation÷number of molecules before dissociation

                         2  =( 1+2y)÷1

                        y=0.5

the degree of dissociation for the given salt is 0.5 or 50%,  in short summary the degree of dissociation can be calculated by the ratio of number of molecules after dissociation to that of number of molecules after dissociation.

Answer 2

Hoff factor = 2 (Given)

For MX2 type salt

MX2 = M²+2X

Number of particles (n) formed after dissociation of MX2 is 3

Degree of dissociation (α) = i-1/n-1

i = vant

hoff factor = 2

Degree of dissociation (α) = 2-1/3-1

                                           = 1/2

                                           = 0.5

Thus, the degree of dissociation of the salt is 0.5