Question

Van't Hoff's factor of 0.1M Ba(NO2)2 solution is 2.74. The degree of dissociation is-

Answer 1

87% B option is correct because we have n(1+2x)

Answer 2

Answer: 0.174

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

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

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

  0.1               0             0

$0.1-\alpha$        $\alpha$          $\alpha$  

Total moles after dissociation=$0.1-\alpha+\alpha+\alpha=0.1+\alpha$   Thus $i=\frac{0.1+\alpha}{0.1}$

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

$\alpha=0.174$