Question

Observed Molar mass of Ca(NO3)2 is 131.2 by elevation of boiling point method. The degree of
dissociation of Ca(NO3)2 is: [Given: Atomic mass of Ca = 40u]
(1) 100%
(2) 75%
(3) 50%
(4) 12.5%​

Answer 1

Answer:

12.5%.

Explanation:

We know that the ionization of Ca(NO3)2 will be Ca(NO3)2 -> Ca2+ + 2NO3- so we will get the value of n to be 3 since on ionization 1 moles of Ca+2 and 2 moles of NO3- is formed.

Thus, we know that the vanthoff factor is having the formulae of observed mass / calculated mass which will be 164/131.2 = 1.25.

So, the formulae of dissociation is i = 1 + α(n - 1) where is the degree of dissociation so 1.25 = 1 + α(3-1) which on solving we will get the value of α to be 0.125 hence the dissociation will be 12.5%.