Question

A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. calculate the osmotic pressure of the solution, $\left( {R = 8.314J{K^{ - 1}}mo{l^{ - 1}}} \right)$.

Answer 1

Hint: We will understand the value of molarity after considering the fact that the solution is decimolar. Then we will convert its unit and dissociate the ferrocyanide and write its equation. Then we will apply the formula for osmotic pressure to get our answer.

Complete step by step solution:

Formula used: ${\pi _{\exp }} = CRT \times$(sum of the number of moles of the reactant and the products).
Given that the solution is a decimolar solution, it is clear that it means that the molarity of the solution is-
Molarity $= 0.1Mol/Lt.$
When we will convert the unit of the above mentioned quantity from $Mol/Lt.$ to $Mol/{m^3}$, we will multiply the given quantity with 1000, we have-
$\Rightarrow 0.1 \times 1000 \\\ \\\ \Rightarrow 100mol/{m^3} \\\$
It can be written as C because it is a concentration term. So-
$\Rightarrow C = 100mol/{m^3}$
Potassium ferrocyanide $= {K_4}Fe{\left( {CN} \right)_6}$
The above compound dissociates as follows-
$\Rightarrow {K_4}Fe{\left( {CN} \right)_6} \to 4{K^ + } + Fe\left( {CN} \right)_6^{4 - }$
Before dissociation, the number of moles of potassium ferrocyanide was 1 and the zero moles of the product. After dissociation, let the reduced amount be $\alpha$, then the number of moles of potassium ferrocyanide were $1 - \alpha$. And the number of moles of ${K^ + }$ were $4\alpha$ and the number of moles of $Fe\left( {CN} \right)_6^{4 - }$ were $\alpha$.
In normal terms, osmotic pressure is written as ${\pi _N} = CRT$.
In experimental terms, it is written as ${\pi _{\exp }} = CRT \times$(sum of the number of moles of the reactant and the products).
$\Rightarrow {\pi _{\exp }} = CRT\left( {1 - \alpha + 4\alpha + \alpha } \right) \\\ \\\ \Rightarrow {\pi _{\exp }} = CRT\left( {1 + 4\alpha } \right) \\\$
Putting the values of temperature, gas constant and the concentration from the question, we have-
$\Rightarrow {\pi _{\exp }} = CRT\left( {1 + 4\alpha } \right) \\\ \\\ \Rightarrow {\pi _{\exp }} = 100 \times 8.314 \times 300 \times \left( {1 + 4\alpha } \right) \\\$
The value of $\alpha$ is given by the question as there is 50% dissociation hence the value is 0.5. putting this value of $\alpha$ in the above equation, we get-
$\Rightarrow {\pi _{\exp }} = 100 \times 8.314 \times 300 \times \left( {1 + 4 \times 0.5} \right) \\\ \\\ \Rightarrow {\pi _{\exp }} = 7.48 \times {10^5}N/{m^2} \\\$
Hence, the osmotic pressure will be $7.48 \times {10^5}N/{m^2}$.

Note: Osmotic pressure can be defined as the minimum pressure needed for solutions to stop solvent molecules from flowing through a semipermeable membrane (osmosis). It is a collective property and depends on the solvent concentration in the solution. With the assistance of the following formula, osmotic pressure can be determined: $\pi = iCRT$. Where, i is the Van’t Hoff factor, R is the universal gas constant, T is the temperature, C is the molar concentration of the solute in the solution and $\pi$ is the symbol for osmotic pressure.