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

Options

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

Answer

B

Explanation

`{:(" "Ba(NO_(3))_(2)hArrBa^(2+)+2NO_(3)),("Initial 0.1 M"),("At. eqm."" "(0.1-x)M " xM 2xM"):}`
`i=((0.1-x)+alpha+2x)/(0.1)=(0.1+2x)/(0.1)=2.74`
`0.1+2x=0.274`
or `x=(0.174)/(2)=0.087`
`%alpha=(x)/(0.1)xx100=(0.087)/(0.1)xx100=87%`