Question

Which have van’t Hoff factors the same as ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$.
A. $A{l_2}{\left( {S{O_4}} \right)_3}$
B. $Mg{\left( {N{O_3}} \right)_2}$
C. $CaC{l_2}$
D. $NaN{O_3}$

Answer 1

van’t Hoff factor gives a deep understanding of the effect of the solutes on the colligative properties of solutions. Colligative properties are those properties that depend upon the number of solute particles (molecules or ions) and not upon the nature of the solute. In the case of non-electrolytes such as sugar, the experimental value of the colligative property agrees with the theoretically calculated value because one mole of a nonelectrolyte gives one mole of particles. However, a solution containing one mole of an electrolyte such as NaCl contains 2 moles of particles (1 mole of $N{a^ + }$ and one mole of $C{l^ - }$). Similarly, 1 mole of $CaC{l_2}$would produce 3 moles of ions in solutions. The particles in the solution will be less when the association takes place. Since the colligative properties of the solution depend upon the number of particles in the given volume, the value observed becomes higher or lower than the calculated values.
Dissociation leads to an increase in the number of particles, hence the colligative properties will show higher values. In the case of an association, the number of solute particles decreases, and consequently, the colligative properties will show abnormally reduced values.
The Van't Hoff factor is denoted by ‘i’.

Complete step by step answer:
Let us calculate the van’t Hoff factor for ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ dissociates into $4{K^ + }$ and ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$
$i = 1 + 4 = 5$
Van’t Hoff factor for ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ is 5.

A.First option is $A{l_2}{\left( {S{O_4}} \right)_3}$ . Let us calculate the van’t Hoff factor for $A{l_2}{\left( {S{O_4}} \right)_3}$.$A{l_2}{\left( {S{O_4}} \right)_3}$ dissociates into $2A{l^{3 + }}$ and $3SO_4^{2 - }$.
$\Rightarrow i = 3 + 2 = 5$
Van’t Hoff factor for $A{l_2}{\left( {S{O_4}} \right)_3}$ is 5.

B.Second option is $Mg{\left( {N{O_3}} \right)_2}$. Let us calculate the van’t Hoff factor for $Mg{\left( {N{O_3}} \right)_2}$.
$Mg{\left( {N{O_3}} \right)_2}$ dissociates into $M{g^{2 + }}$ and $2NO_3^ -$
$\Rightarrow i = 2 + 1 = 3$
Van’t Hoff factor for $Mg{\left( {N{O_3}} \right)_2}$ is 3.

C.Third option is $CaC{l_2}$. Let us calculate the van’t Hoff factor for $CaC{l_2}$.
$CaC{l_2}$ dissociates into $C{a^{2 + }}$ and $2C{l^ - }$
$\Rightarrow i = 2 + 1 = 3$
Van’t Hoff factor for $CaC{l_2}$ is 3.

4.Fourth option is $NaN{O_3}$ . Let us calculate the van’t Hoff factor for $NaN{O_3}$.
$NaN{O_3}$ dissociates into $N{a^ + }$and $NO_3^ -$
$\Rightarrow i = 1 + 1 = 2$
Van’t Hoff factor for $NaN{O_3}$ is 2.

After discussing we can conclude that van’t Hoff factor of $A{l_2}{\left( {S{O_4}} \right)_3}$ is same as van’t Hoff factor of ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ which is equal to 5.

So, the correct answer is Option A .

Note: Degree of Association: - The degree of association is defined as the fraction of the total number of molecules which combine to form associated molecules viz. dimers, trimmers, etc.

Degree of dissociation: - The degree of dissociation is defined as the fraction of the total number of molecules which dissociate, i.e. break into simpler molecules or ions.