Question

If the extent of dissociation of both $KCl$ and $BaC{l}_2$ of the same concentration with an identical value of $\alpha$ is 0.95, what is the ratio of their van' t Hoff factors:

• A. 1.95 : 2.9
• B. 1.89 : 1
• C. 1 : 2.21
• D. 2.11 : 1

Answer 1

The correct option is A 1.95 : 2.9

$\begin{array}{ccccc}KCl& \to & K^{+}& +& C{l}^{-}\\ (1-\alpha )& & \alpha & & \alpha \end{array}$
$i=1-\alpha +\alpha +\alpha =1+\alpha =1+0.95=1.95$

$\begin{array}{ccccc}BaC{l}_2& \to & B{a}^{+2}& +& 2C{l}^{-}\\ (1-\alpha )& & \alpha & & 2\alpha \end{array}$
$i=1-\alpha +\alpha +2\alpha =1+2\alpha =1+2\left(0.95\right)$
$=2.9$
Then the ratio of van' t Hoff factors for $KCl$ and $BaC{l}_2=1.95:2.9$