Question

If $0.1\left( \text{M} \right)$solutions of ${{\text{K}}_{\text{4}}}\left[ \text{Fe}{{\left( \text{CN} \right)}_{\text{6}}} \right]$is prepared at 300$\text{K}$then its density= $1.2\text{g/mL}$. If solute is 50% dissociated, calculate$\text{ }\!\!\Delta\!\!\text{ P}$of solution if $\text{P}$of pure water= $\text{25 mm of Hg}$$\left( \text{K=39, Fe=56} \right)$

Answer 1

Hint: This question can be answered on the basis of the colligative property, Relative Lowering of Vapour Pressure. The colligative properties of solutions are those that depend on the number of the solute particles present in the medium, irrespective of their nature. The lowering of vapour pressure of solution is one such property.

Formula used: \[\dfrac{{{\text{P}}^{\text{0}}}\text{-}{{\text{P}}^{\text{s}}}}{{{\text{P}}^{0}}}\text{=i}\dfrac{\text{w}}{\text{m}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{M}}{\text{W}}\]
where, \[{{\text{P}}^{0}}\]is the vapour pressure of the solvent, \[{{\text{P}}^{\text{s}}}\]is the vapour pressure of the solution, ‘i’ is called the Van’t Hoff Factor, w is the mass of the solute and m is its molecular weight, while M is the molecular weight of the solvent and W is the mass of the solvent.
Degree of dissociation, $\text{ }\!\!\alpha\!\!\text{ }=\dfrac{\text{i}-1}{\text{n}-1}$
where, $\text{ }\!\!\alpha\!\!\text{ }$is the degree of dissociation of electrolytes and n is the number of particles in the solution.

Complete Step-wise solution:
 From the question we get that, 50% dissociation of the solution is there. Therefore, $\text{ }\!\!\alpha\!\!\text{ }$=$0.5$ and the number of particles in the medium are five.
Putting these values in the above equation for $\text{ }\!\!\alpha\!\!\text{ }$ we get,
$\Rightarrow \text{i}=0.5(5-1)+1=3$
So the Van’t Hoff Factor = 3.
Concentration of the solution=$0.1\left( \text{M} \right)$ which means that there are $0.1$ moles of solute per litre of the solution.
Molecular mass of the solute, m=$\left[ \left( 39\times 4 \right)+56+6\left( 12+14 \right) \right]=368$g
Mass of solute in solution, w=$36.8$g for $0.1\left( \text{M} \right)$ solution.
Mass of the solvent=mass of the solution-mass of the solute=\[1200-36.8=1163.2\]g
No. of moles of the solvent in the solution:
\[{{\text{x}}_{\text{s}}}=\dfrac{1163.2}{18}=64.62\]moles.
So the mole-fraction of the solute in the solution = $\dfrac{0.1}{0.1+64.62}=0.00154$
Modifying the above formula for relative lowering of vapour pressure we get, \[\dfrac{\text{ }\!\!\Delta\!\!\text{ }{{\text{p}}^{\text{s}}}}{{{\text{p}}^{\text{0}}}}\text{=i }\!\!\times\!\!\text{ }{{\text{x}}_{\text{s}}}\]
Therefore, the relative lowering in the vapour pressure of water due to the addition of${{\text{K}}_{\text{4}}}\left[ \text{Fe}{{\left( \text{CN} \right)}_{\text{6}}} \right]$ is $\Delta {{\text{P}}^{\text{s}}} = 3 \times 0.00154 \times 25$ = \[0.1155\]mm of Hg.

Notes: The solutes that are electrolyte dissolve in water to dissociate into ions. These ions are responsible for the change in the colligative properties of the solution as the properties are totally dependent on the number of the solute particles present in the medium and not on the nature of the solute. Similarly, there are solutes that associate too, such as acetic acid. This association is due to the formation of intra-molecular hydrogen bond that also affects the properties of the solution.