Decimolar solution of potassium ferrocyanide is 50%dissociated at 300 k , osmotic pressure ?
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Dissociation reaction of potassium ferrocyanide is
K₄[Fe(CN)₆] ⇔4K⁺ + [Fe(CN)₆]⁴⁻
So, vn Hoff's factor , i = 1 - α + 4α + α = 1 + 4α
But α is given 50%
So, α = 0.5
Now, i = 1 + 4α = 1 + 4 × 0.5 = 1 + 2 = 3
Given, concentration , C = decimolar = 0.1M
Temperature , T = 300K
R = 0.082 atm.L/K/mol
Now, Osmotic pressure , π = iCRT
= 3 × 0.1 × 0.082 × 300
= 900 × 0.0082
= 7.38 atm
Hence, Osmotic pressure = 7.38 atm
K₄[Fe(CN)₆] ⇔4K⁺ + [Fe(CN)₆]⁴⁻
So, vn Hoff's factor , i = 1 - α + 4α + α = 1 + 4α
But α is given 50%
So, α = 0.5
Now, i = 1 + 4α = 1 + 4 × 0.5 = 1 + 2 = 3
Given, concentration , C = decimolar = 0.1M
Temperature , T = 300K
R = 0.082 atm.L/K/mol
Now, Osmotic pressure , π = iCRT
= 3 × 0.1 × 0.082 × 300
= 900 × 0.0082
= 7.38 atm
Hence, Osmotic pressure = 7.38 atm
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24
7.38
hope this helps.....
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