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

Options

A. `91.3%`
B. 0.87
C. 1
D. 0.74

Answer

B

Explanation

`{:(,":",Ba(NO_(3))_(2)hArr ,Ba^(2+) + 2NO_(3)^(-)),(,"Initial", " " 1 mol,"- -"),(,"After dissociation"," "1-alpha, alpha" "2alpha):}`
Total moles `= 1 - alpha + alpha + 2alpha = 1 + 2 alpha`
` i= 1 + 2alpha or alpha`
` = (i-1)/(2) = (2.74 - 1)/(2) = 0.87 or 87%`