Question

Which of the following salt has the same value of Van’t Hoff factor ‘i’ as that of ${{K}_{3}}[Fe{{(CN)}_{6}}]$ ?
(A) $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$
(B) NaCl
(C) $N{{a}_{2}}S{{O}_{4}}$
(D) $Al{{(N{{O}_{3}})}_{3}}$

Answer 1

Hint: The Van’t Hoff factor gave insight into the effect of solutes on the colligative properties of solutions. This factor describes the extent to which a substance associates or dissociates in a solution. For example, when an ionic compound dissolves in a solution of water, the value of ‘i’ is equal to the total number of ions present in one formula of the unit of the substance. However, the value of ‘i’ is 1 for non-electrolytic substances dissolved in water.

Complete step by step solution:
The Van’t Hoff factor is defined as the ratio of the concentration of particles formed when a substance dissolved to the concentration of the substance. The factors effects on Van’t Hoff factor are association or dissociation of a solute on the solution and their colligative properties. Association means the joining of two or more particles to form one compound or one entity. For example, when carboxylic acid dissolves in benzene, the association of two particles is the dimerization of carboxylic acid. Dissociation means the splitting of a molecule into multiple ionic entities. For example, when NaCl dissolves in water, then it dissociates into a sodium ion and chloride ion.
Given, ${{K}_{3}}[Fe{{(CN)}_{6}}]$ will dissociate into four particles as shown below,
${{K}_{3}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+Fe{{(CN)}_{6}}^{3-}$
Similarly, $Al{{(N{{O}_{3}})}_{3}}$ also dissociates into 4 particles as shown below,
$Al{{(N{{O}_{3}})}_{3}}\to A{{l}^{3+}}+3N{{O}_{3}}^{-}$
For NaCl, $A{{l}_{2}}{{(S{{O}_{4}})}_{3}},N{{a}_{2}}S{{O}_{4}}$ are dissociates into 2 particles, 5 particles, and 3 particles respectively as shown below.
$\begin{align}
& NaCl\to N{{a}^{+}}+C{{l}^{-}} \\
& N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+S{{O}_{4}}^{-2} \\
& A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 2A{{l}^{+}}+3S{{O}_{4}}^{2-} \\
\end{align}$
Hence, $Al{{(N{{O}_{3}})}_{3}}$ has the same value of Van’t Hoff factor ‘i’ as that of ${{K}_{3}}[Fe{{(CN)}_{6}}]$, because the dissociation particles are the same.

So, the correct answer is option D.

Note: Van’t Hoff factor for dissociation, the value will be greater than 1 in absence of association, and quantity of solute increases, colligative properties increases, the molar mass of solute decreases. Van’t Hoff factor for the association, the value will be smaller than 1, and both quantity of solute, colligative properties decreases, the molar mass of solute decreases.