Question

Which one of the following electrolytes has the same value of van’t Hoff factor (i) as that of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ (if all are 100% ionized)?
(a)- $Al{{(N{{O}_{3}})}_{3}}$
(b)- ${{K}_{4}}[Fe{{(CN)}_{6}}]$
(c)- ${{K}_{2}}S{{O}_{4}}$
(d)- ${{K}_{3}}[Fe{{(CN)}_{6}}]$

Answer 1

Hint: Van’t Hoff factor can be calculated by the reaction of the compound when it is completely ionized in the solution. It can be calculated as the ratio of the number of solute particles present in solution to the theoretical number of solute particles due to the solution of a nonelectrolyte.

Complete answer:
The Van't Hoff factor can be calculated by the reaction of the compound when it is completely ionized in the solution. It can be calculated as the ratio of number of solute particles present in solution to the theoretical number of solute particles due to solution of non electrolyte.
$i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
So, the dissociation of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ will be:
$A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 2A{{l}^{3+}}+3SO_{4}^{2-}$
So, 1 mole of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ produces 2 moles of aluminium ions and 3 moles of sulfate ions (total 5 moles).
So, $i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
$i=\dfrac{5}{1}=5$
The (i) of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$is 5.
Now for $Al{{(N{{O}_{3}})}_{3}}$, the reaction will be:
$Al{{(N{{O}_{3}})}_{3}}\to A{{l}^{3+}}+3NO_{3}^{-}$
So, 1 mole of $Al{{(N{{O}_{3}})}_{3}}$produces 1 mole of aluminium and 3 moles of nitrate ions (total 4 moles).
So, $i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
$i=\dfrac{4}{1}=4$
The (i) of$Al{{(N{{O}_{3}})}_{3}}$ is 4.
Now, for ${{K}_{4}}[Fe{{(CN)}_{6}}]$, the reaction will be:
${{K}_{4}}[Fe{{(CN)}_{6}}]\to 4{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{4-}}$
So, 1 mole of ${{K}_{4}}[Fe{{(CN)}_{6}}]$ produces 4 moles of potassium ion and 1 mole of ferricyanide ion (total 5 moles).
So, $i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
$i=\dfrac{5}{1}=5$
The (i) of ${{K}_{4}}[Fe{{(CN)}_{6}}]$ is 5.
Now for ${{K}_{2}}S{{O}_{4}}$, the reaction will be:
${{K}_{2}}S{{O}_{4}}\to 2{{K}^{+}}+SO_{4}^{2-}$
So, 1 mole of ${{K}_{2}}S{{O}_{4}}$produces 2 moles of potassium ion and 1 mole of sulfate ion (total 3 moles).
So, $i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
$i=\dfrac{3}{1}=3$
The (i) of ${{K}_{2}}S{{O}_{4}}$ is 3.
Now for ${{K}_{3}}[Fe{{(CN)}_{6}}]$, the reaction will be:
${{K}_{3}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}}$
So, 1 mole of ${{K}_{4}}[Fe{{(CN)}_{6}}]$ produces 3 moles of potassium ion and 1 mole of ferricyanide ion (total 4 moles).
So, $i=\dfrac{n\,(observed)}{n\text{ }(theoretical)}$
$i=\dfrac{4}{1}=4$
The (i) of ${{K}_{3}}[Fe{{(CN)}_{6}}]$ is 5.

Therefore, the correct answer is an option (b)- ${{K}_{4}}[Fe{{(CN)}_{6}}]$.

Note:
The van’t Hoff factor is not always the same as the calculated one. There are two values van’t Hoff factor, i.e., observed and experimental. The experimental value can be calculated from any formula of colligative property.