Question
A deci-molar solution of K4 [Fe(CN)6] is 50% dissociated at 300 K. Calculate the osmotic pressure of solution (R = 0.082)/L atm K–1)?
Solution
Answer:
The given complex is ionsiable and it dissociates as follow
(1-a) 4a a
Where a is the degree of dissociation of a complex.
intial moles of complex =1
moles after dissociation of the complex = 1-a + 4a+a = 1+4a
The given complex is ionsiable and it dissociates as follow
(1-a) 4a a
Where a is the degree of dissociation of a complex.
intial 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
Pobserve/ Pnormal= (1+4a)/1
Dissociation takes place only 50% a= 0.50
P/2.462 = 1+(4 x 0.50)
p/2.462 = 1+2
p/2.462=3
p= 2.462x 3
p=7.386 atm
Pobserve/ Pnormal= (1+4a)/1
Dissociation takes place only 50% a= 0.50
P/2.462 = 1+(4 x 0.50)
p/2.462 = 1+2
p/2.462=3
p= 2.462x 3
p=7.386 atm
