The van't Hoff factor for `0.1 M Ba(NO_(3))_(2)` solution is `2.74` . The degree of dissociation is :

Options

A. 0.87
B. 0.74
C. 0.91
D. 87

Answer

A

Explanation

Given,
Molarity = 0.1 M
vant Hoff factor ( i) = 2.74
Since, `igt1`, it means solute is undergoing dIssociation.
`Ba(NO_(3))_(2)hArrBa^(2+)+2NO_(3)^(-)`
Number of particles dissociated (n) = 3
Now, `alpha` (degree of dissociation)`=((i-1))/((n-1))`
`=((2.74-1))/((3-1))`
= 0.87