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Question

a decimolar solution of potassium ferrocyanide is 5% dissociated at 300 Kelvin calculate Osmotic pressure of solution

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Solution

When the complex is not dissociated then we can use the Van 't Hoff equation to calculate the normal osmotic pressure.

P = c*R*T

Where, P = Osmotic pressure,

c = concentration

R = gas constant

T = temperature in Kelvin

Given values, c= 0.1 molar , T = 300 K

P = 0.1 * 0.0821 * 300

P = 2.462 atm

However, the given complex is ionisable and it dissociates as follows....

K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4-

1-a 4a a

where, a is the degree of dissociation of the complex.

Initial moles of complex = 1

Moles after dissociation of the complex= 1 - a + 4a +a = 1 + 4a

Since, osmotic pressure is directly proportional to the number of moles, hence

P(observed)/P(normal) = (1+4a)/1

Dissociation takes place only 5%. so, a = 0.050

P/2 x 462 = 1+(4*0.050)

P/2 x 462 = 1 + 0.2

P/2 x 462 = 1.2

P = 2.462 x 1.2

P = 2.9544 atm


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