Calculate the osmotic pressure of the solution if a decimolar solution of ferrocyanide is $50\% $dissociated at $300K$ . (Given: $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$).
Answer
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Hint: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: $\pi = iCRT$
The dissociation of one molecule of ${K_4}\left [ {Fe{{\left( {CN} \right)}_6}} \right]$ gives five ions.
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftharpoons 4{K^ + } + {\left[ {Fe{{(CN)}_6}} \right]^{4 - }}$
Since the solution is $50\% $ dissociated, the Van't Hoff’s factor is as follows:
$i = [1+(n-1) \alpha] $
$\alpha = \ degree\, of\, dissociation$ = 0.5.
The solution is $50\% $ dissociated. Thus $i = \dfrac{{1 + 4(0.5)}}{1} = 3$ = $i = 3$
Substituting values $T = 300K$,$C = 0.1$, $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$ and $i = 3$ in the above expression: $\pi = iCRT$
$\pi = 3 \times 0.1 \times 8.314 \times 300 = 748.26J{L^{ - 1}}$
Now the conversion of units:
$8.314J = 0.0821Latm$
Thus, $1J{L^{ - 1}} = 0.00987atm$
$748.26J{L^{ - 1}} = 7.385atm$
$1atm = 1.01325 \times {10^5}N{m^{ - 2}}$
$\therefore 7.385atm = 7.482 \times {10^5}N{m^{ - 2}}$
Hence the osmotic pressure of solution is $7.482 \times {10^5}N{m^{ - 2}}$.
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.
Complete step-by-step solution:The expression for the osmotic pressure of solution is: $\pi = iCRT$
The dissociation of one molecule of ${K_4}\left [ {Fe{{\left( {CN} \right)}_6}} \right]$ gives five ions.
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftharpoons 4{K^ + } + {\left[ {Fe{{(CN)}_6}} \right]^{4 - }}$
Since the solution is $50\% $ dissociated, the Van't Hoff’s factor is as follows:
$i = [1+(n-1) \alpha] $
$\alpha = \ degree\, of\, dissociation$ = 0.5.
The solution is $50\% $ dissociated. Thus $i = \dfrac{{1 + 4(0.5)}}{1} = 3$ = $i = 3$
Substituting values $T = 300K$,$C = 0.1$, $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$ and $i = 3$ in the above expression: $\pi = iCRT$
$\pi = 3 \times 0.1 \times 8.314 \times 300 = 748.26J{L^{ - 1}}$
Now the conversion of units:
$8.314J = 0.0821Latm$
Thus, $1J{L^{ - 1}} = 0.00987atm$
$748.26J{L^{ - 1}} = 7.385atm$
$1atm = 1.01325 \times {10^5}N{m^{ - 2}}$
$\therefore 7.385atm = 7.482 \times {10^5}N{m^{ - 2}}$
Hence the osmotic pressure of solution is $7.482 \times {10^5}N{m^{ - 2}}$.
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.
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