The van't Hof factor for 0.1 m `Ba(NO_(3))_(2)` solution is 2.74 percentage of dissociation is …..

Options

A. `91.3%`
B. `100%`
C. `87%`
D. `74%`

Answer

C

Explanation

`Ba(NO_(3))_(2)` ionises as …….
`{:(Ba(NO_(3))_(2),to,Ba^(2+),+,2NO_(3)^(-)),("No of ions formed =",,(1),+,(2)):}`
`therefore n = 3` and `i=2.74`
Degree of dissociation,
`alpha = (i-1)/(n-1)`
`= (2.74-1)/(3-1)`
`= (1.74)/(2)=0.87=87%`