Question

# A decimolar solution of K4[Fe(CN)6] is 50% dissociated at 300 K. Calculate the osmotic pressure of solution. (R= 0.0821 litre atm/K mol)

A
2.3 atm
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B
7.39 atm
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C
3.25 atm
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D
None of these
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Solution

## The correct option is B 7.39 atmWhen the complex is not dissociated we can use Van't Haff equation to calculate normal osmotic pressure,p=C×R×T⟹P=0.1×0.0821×300=2.462atmThe given complex dissociates and it dissociates as follows,K4[Fe(CN)6]→4K++[Fe(CN)6]4− (1−a) 4a aWhere a is degree of dissociation of the complex assuming initially 1 mole was present.Moles after dissociation of the complex=1−a+4a+a=1+4a,Since osmotic pressure is directly proportional to number of moles,pobservedpnormalen=1+4a1Given dissociation is 50% and hence a=0.5So,p2.462=1+4×0.5⟹p=7.386atm

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