Question

The van’t hoff factor is 3 for
A. ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ with $50\%$ dissociation
B. ${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}}$
C. ${\rm{CaC}}{{\rm{l}}_2}$ with $80\%$ dissociation
D. ${{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right]$ with $50\%$ dissociation

Answer 1

The degree of dissociation is the fraction of the original solute molecules that have dissociated. For dissociation in the absence of association, the van 't Hoff factor is ${\rm{I > 1}}$. For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
The extent to which a substance associates or dissociates in a solution is described by the Van’t Hoff factor. For example, when a non-electrolytic substance is dissolved in water, the value of i is generally 1. However, when an ionic compound forms a solution in water, the value of i is equal to the total number of ions present in one formula unit of the substance.
For example, the Van’t Hoff factor of ${\rm{CaC}}{{\rm{l}}_{\rm{2}}}$ is ideally 3, since it dissociates into one ${\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}$ ion and two ${\rm{C}}{{\rm{l}}^{\rm{-}}}$ ions. However, some of these ions associate with each other in the solution, leading to a decrease in the total number of particles in the solution.

Complete step by step answer:
A. ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ with $50\%$ dissociation
${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {\rm{2N}}{{\rm{a}}^{\rm{ + }}}{\rm{ + SO}}_{\rm{4}}^{{\rm{2 - }}} \\\ {\rm{1 - 0}}{\rm{.5\;\;\;\;\;\;\;2 \times 0}}{\rm{.5\;\;\;0}}{\rm{.5}}$
Van't Hoff factor ${\rm{ = 1 - 0}}{\rm{.5 + 2 \times 0}}{\rm{.5 + 0}}{\rm{.5 = 2}}$

B. ${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}}$
${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}} \to {\rm{2A}}{{\rm{l}}^{{\rm{3 + }}}}{\rm{ + SO}}_{\rm{4}}^{{\rm{2 - }}}$
${\rm{I = 5}}$

C. ${\rm{CaC}}{{\rm{l}}_2}$ with $80\%$ dissociation
${\rm{CaC}}{{\rm{l}}_{\rm{2}}} \to {\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}{\rm{ + 2C}}{{\rm{l}}^{\rm{ - }}} \\\ {\rm{1 - 0}}{\rm{.8\;\;\;\;\;0}}{\rm{.8\;\;\;\;\;\;\;2 \times 0}}{\rm{.8}}$
Van't Hoff factor ${\rm{ = 0}}{\rm{.2 + 0}}{\rm{.8 + 1}}{\rm{.6 = 2}}{\rm{.6}}$

D. ${{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right]$ with $50\%$ dissociation
${{\rm{K}}_{\rm{4}}}{\rm{[Fe(CN}}{{\rm{)}}_{\rm{6}}}{\rm{]}} \to {\rm{4}}{{\rm{K}}^{\rm{ + }}}{\rm{ + Fe(CN)}}_{\rm{6}}^{{\rm{4 - }}} \\\ {\rm{1 - 0}}{\rm{.5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4 \times 0}}{\rm{.5\;\;\;\;\;0}}{\rm{.5}}$
Van't Hoff factor ${\rm{ = 0}}{\rm{.5 + 2 + 0}}{\rm{.5 = 3}}$

**Option D is correct

Note:**
The extent to which a substance associates or dissociates in a solution is described by the Van’t Hoff factor.Dissociation refers to the splitting of a molecule into multiple ionic entities.
For example, sodium chloride ${\rm{NaCl}}$ dissociates into ${\rm{N}}{{\rm{a}}^{\rm{ + }}}{\rm{\;and\;\ C}}{{\rm{l}}^{\rm{ - }}}$ ions when dissolved in water.