Question

If $\dfrac{{{M}}}{{{{10}}}}$ solution of potassium ferrocyanide is $46\%$ dissociated at ${{18^\circ C}}$. What will be its osmotic pressure?
(A) 2.84 atm
(B) 4.892 atm
(C) 8.612 atm
(D) 6.785 atm

Answer 1

The osmotic pressure can be determined by using van’t Hoff equation for the dilute solutions as:
${{\pi V}} = {{nRT}}$or ${{\pi }} = \dfrac{{{n}}}{{{V}}}{{RT}}$
${{\pi }} = {{CRT}}$ $......\left( 1 \right)$
Where, ${{\pi }}$ is osmotic pressure, C is the concentration of the solution, R is the gas constant and T is the temperature.

Complete step by step answer:
Osmotic pressure can be defined as the minimum excess pressure that has to be applied to the solution to prevent the entry of solvent via a semipermeable membrane. It is used in desalination of seawater.
Equation (1) is valid for non-electrolytic solution but the given problem is of an electrolytic solution to give five ions. Then, the above formula will be given as:
${{\pi }} = {{iCRT}}$ $......\left( 2 \right)$
Here, i is the van’t Hoff factor.
Now, potassium ferrocyanide dissociates to give the following reaction as:
${{{K}}_{{4}}}\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_{{6}}}} \right] \to {\left[ {{{Fe}}{{\left( {{{CN}}} \right)}_{{6}}}} \right]^{4 - }} + {{4}}{{{K}}^ + }$ $\left( {{{n}} = 5} \right)$
As potassium ferrocyanide is 46% dissociated than the value of van’t Hoff factor (i) can be determined as:
${{\alpha }} = 46\% = \dfrac{{46}}{{100}} = 0.46$
${{\alpha }} = \dfrac{{{{i}} - 1}}{{{{n}} - 1}}$
$0.46 = \dfrac{{{{i}} - 1}}{{5 - 1}}$
i = 2.84
Now, substitute the values in equation (2), we will get the osmotic pressure as:
${{\pi }} = 2.84 \times \dfrac{1}{{10}} \times 0.0821 \times 291$
${{\pi }} = {{6}}{{.785}}\;{{atm}}$

So, the correct answer is Option D.

Note: The number of methods is available for the determination of osmotic pressure. The best method is Berkeley and Hartley’s method, which is based on applying sufficient pressure to the solution to prevent the entry of solvent via semi permeable membrane.