Question

the van't Hoffs factor of 0.1M Ba(NO2)2 solution is 2.74 . the degree of dissociation is
A. 91.3%
B. 87%
C. 100%
D. 74%

Answer 1

degree of dissociation is 87%

Answer 2

Answer: B. 87%

Explanation: Van't hoff's factor is the ratio of observed colligative property to the calculated colligative property.

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

Colligative property is the property which depends on the amount of solute.

$Ba(NO_2)_2\rightarrow Ba^{2+}+2NO_2^{-}$

initial:         0.1            0             0

after dissociation: $0.1-\alpha$        $\alpha$          $2\alpha$  

where$\alpha$= degree of dissociation

$i=\frac{\text {observed moles}}{\text {Calculated moles}}$

Observed moles after dissociation =$0.1-\alpha+\alpha+2\alpha=0.1+2\alpha$

$2.74=\frac{0.1+2\alpha}{0.1}$

$\alpha=0.087=87\%$