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The van't Hoff factor for 0.1 M Ba(NO(3)...

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

A

`91.3 %`

B

`87%`

C

`100%`

D

`74%`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ba(NO_(3))_(2)to Ba^(+2)+2NO_(3)^(-)`
`therefore` Number of ions produced = 3 = n
`i=2.74`
`therefore alpha = (i-1)/(n-1)xx 100`
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