Question

A solution contains $\text{3}\text{.22 g}$ of $\text{HCl}{{\text{O}}_{\text{2}}}$ in $\text{47}\text{.0 g}$ of water. The freezing point of the solution is $\text{271}\text{.10 K}$.
Calculate the fraction of $\text{HCl}{{\text{O}}_{\text{2}}}$ that undergoes dissociation to ${{\text{H}}^{\text{+}}}$and $\text{Cl}{{\text{O}}_{\text{2}}}^{-}$. Given the ${{\text{K}}_{\text{f}}}$(water) = $\text{1}\text{.86 K Kg mo}{{\text{l}}^{\text{-1}}}$

Answer 1

Hint: When a non-volatile solute is added to a solvent then the freezing point of the resulting solution is lowered and this phenomenon is called the “Depression in freezing point” and this is a colligative property of the solution which is dependent on the moles of the solute present in the solution.

Complete step by step answer:
Given that, the mass of $\text{HCl}{{\text{O}}_{\text{2}}}$= $\text{3}\text{.22 g}$
The mass of water required to form the solution = $\text{47}\text{.0 g}$= $\text{0}\text{.047 Kg}$
The molecular weight of $\text{HCl}{{\text{O}}_{\text{2}}}$= $\left[ 1+35.5+\left( 16\times 2 \right) \right]=68.5$$\text{g/ mol}$
Therefore number of moles of chlorous acid present in the solution = \[\dfrac{\text{3}\text{.22}}{\text{68}\text{.5}}\text{ = 0}\text{.047 moles}\]
Therefore the molality of the solution which is the number of moles of the solute present per Kg of the solvent = $\dfrac{\text{0}\text{.047}}{\text{0}\text{.047}}\text{=1 (m)}$
Therefore the depression in the freezing point of the solution can be given by,
$\text{ }\Delta\text{ }{{\text{T}}_{\text{f}}}=\text{i}\times {{\text{k}}_{\text{f}}}\text{ }\times\text{ m}$, where “i” is the van’t Hoff factor.
Given that, ${{\text{K}}_{\text{f}}}$ (water) = $\text{1}\text{.86 K Kg mo}{{\text{l}}^{\text{-1}}}$,
The freezing point of water = $\text{273}\text{.15 K}$
The freezing point of the solution = $\text{271}\text{.10 K}$
Therefore the depression in the freezing point = $\left( \text{273}\text{.15 - 271}\text{.10} \right)\text{ = 2}\text{.05 K}$
Putting the values in the above equation we get,
$\text{i}=\dfrac{2.05}{1.86\times 1}=1.102$
The fraction of chlorous acid that undergoes dissociation, $\text{ }\alpha\text{ =}\dfrac{\text{i -1}}{\text{n -1}}$
Where n is the number of the species to which the chlorous acid dissociates to:
$\text{ }\alpha\text{ =}\dfrac{\text{1}\text{.102 -1}}{\text{2 -1}}=0.102$
The fraction of $\text{HCl}{{\text{O}}_{\text{2}}}$that undergoes dissociation to ${{\text{H}}^{\text{+}}}$and $\text{Cl}{{\text{O}}_{\text{2}}}^{-}$ is $0.102$.

Note:
The van't Hoff factor is defined as the ratio of the actual concentration of the particles produced when the substance is dissolved and the concentration of the substance as calculated from its mass. The property offers insights on the effect of the solute on the colligative properties of the solution.