Question

The value of Van't hoff factor 'i' for aqueous KCl solution is close to 2, while the value of 'i' for ethanoic acid in benzene is nearly 0.5. Give reason

Answer 1

Explanations:- Van't hoff factor"i" for KCl is close to 2 since it's ionic compound and ionic compounds ionization is high. KCl breaks to give potassium and chloride ions as:

$KCl\rightarrow K^++Cl^-$

Theoretically one mol of KCl gives one mol of potassium ion and one mol of chloride ion, the total is 2. In actual nothing is 100% and so the actual Van't hoff factor for KCl is close to 2.

Ethanoic acid is not a strong acid, it is a weak acid, and so it's ionization is very less and due to this reason it's Van't hoff factor is also low like 0.5.

So, more is the ionization, higher is the value of Van't hoff factor and vice versa.