The Van't Hoff factor for 0.1 M `CaCl_(2)` solution is 2.74. The degree of dissociation is

Options

A. `61%`
B. 0.87
C. 1
D. 0.54

Answer

B

Explanation

`CaCl_(2) hArr Ca^(2+) +2Cl^(-)`
Initial 1 mole after dissciation ` 1- alpha ` `alpha ` `2alpha`
Total ` = 1 + 2 alpha `
`:. i= 1+ 2alpha `
or `alpha = ( i-1)/( 2) = ( 2.74-1)/( 2) = 0.87 =87%`