Question

the theoretical molecular mass and observed molecular mass of an ionic compound AB are 58.2 and 30 respectively. calculate the van't hoff factor and degree of dissociation​

Answer 1

Answer: vant hoff factor = 1.94

degree of dissociation = 0.94

Explanation:

Vant hoff factor is the ratio of observed colligative property to the calculated colligative property.

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

$i=\frac{\text {calculated molar mass}}{\text {observed Molar mass}}$

$i=\frac{58.2}{30}=1.94$

Thus vant hoff factor will be 1.94.

$AB\rightarrow A^{n+}+B^{n-}$

Degree of dissociation $\alpha$ is defined as the fraction of total molecules which dissociate into ions.

$\alpha=\frac{i-1}{n-1}$

n = number of particles in solution

$\alpha=\frac{i-1}{2-1}$

$\alpha=0.94$

Thus degree of  dissociation is 0.94.