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Question: Calculate the osmotic pressure of the solution if a decimolar solution of ferrocyanide is \(50\% \)d...

Calculate the osmotic pressure of the solution if a decimolar solution of ferrocyanide is 50%50\% dissociated at 300K300K . (Given: R=8.314JK1mol1R = 8.314J{K^{ - 1}}mo{l^{ - 1}}).

Explanation

Solution

Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane. Osmosis occurs when two solutions containing different concentrations of solute are separated by a selectively permeable membrane. Solvent molecules pass preferentially through the membrane from the low-concentration solution to the solution with higher solute concentration. The transfer of solvent molecules will continue until equilibrium is attained.

Complete step-by-step solution: The expression for the osmotic pressure of solution is: π=iCRT\pi = iCRT
The dissociation of one molecule of K4[Fe(CN)6]{K_4}\left [ {Fe{{\left( {CN} \right)}_6}} \right] gives five ions.
K4[Fe(CN)6]4K++[Fe(CN)6]4{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftharpoons 4{K^ + } + {\left[ {Fe{{(CN)}_6}} \right]^{4 - }}
Since the solution is 50%50\% dissociated, the Van't Hoff’s factor is as follows:
i=[1+(n1)α]i = [1+(n-1) \alpha]
α= degreeofdissociation\alpha = \ degree\, of\, dissociation = 0.5.
The solution is 50%50\% dissociated. Thus i=1+4(0.5)1=3i = \dfrac{{1 + 4(0.5)}}{1} = 3 = i=3i = 3
Substituting values T=300KT = 300K,C=0.1C = 0.1, R=8.314JK1mol1R = 8.314J{K^{ - 1}}mo{l^{ - 1}} and i=3i = 3 in the above expression: π=iCRT\pi = iCRT
π=3×0.1×8.314×300=748.26JL1\pi = 3 \times 0.1 \times 8.314 \times 300 = 748.26J{L^{ - 1}}
Now the conversion of units:
8.314J=0.0821Latm8.314J = 0.0821Latm
Thus, 1JL1=0.00987atm1J{L^{ - 1}} = 0.00987atm
748.26JL1=7.385atm748.26J{L^{ - 1}} = 7.385atm
1atm=1.01325×105Nm21atm = 1.01325 \times {10^5}N{m^{ - 2}}
7.385atm=7.482×105Nm2\therefore 7.385atm = 7.482 \times {10^5}N{m^{ - 2}}

Hence the osmotic pressure of solution is 7.482×105Nm27.482 \times {10^5}N{m^{ - 2}}.

Note: The van 't Hoff factor i, is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering in vapor pressure, boiling-point elevation and freezing-point depression. The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.