Question

A decimolar solution potassium ferrocyanide is 50% dissociated at 300 K. The osmotic pressure of solution is (R = 8.314 J K$^{-1}$ mol$^{-1}$): 
 

• A. 7.48 atm
• B. 4.99 atm
• C. 3.74 atm
• D. 6.23 atm

Hint:

Osmotic pressure depends on van't Hoff factor, which accounts for the number of particles in solution.

Answer 1

The Correct Option is A

Step 1: Degree of Dissociation (\(\alpha\)) and Van't Hoff Factor (\(i\))
The reaction for dissociation of K$_4$[Fe(CN)$_6$] is: \[ K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-} \] Since the dissociation is 50\% (\(\alpha = 0.5\)), the total number of particles: \[ i = 1 + 4\alpha = 1 + 4(0.5) = 3 \] Step 2: Applying the Osmotic Pressure Formula
\[ \pi = iCRT \] Substituting values: \[ \pi = (3)(0.1)(8.314)(300) \] \[ \pi = 7.48 { atm} \] Thus, the correct answer is (A).

Answer 2

Step 1: Understand the problem
We are given a 0.1 M (decimolar) solution of potassium ferrocyanide which is 50% dissociated at 300 K.
We need to calculate the osmotic pressure (π) of the solution.

Step 2: Write down the formula for osmotic pressure
π = i × M × R × T
where,
i = van’t Hoff factor (effective number of particles),
M = molarity,
R = gas constant (8.314 J K^-1 mol^-1),
T = temperature in Kelvin.

Step 3: Determine van’t Hoff factor (i)
Potassium ferrocyanide, K₄[Fe(CN)₆], dissociates as:
K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻
Total particles if fully dissociated = 5.
Since it is 50% dissociated, i = 1 + α(n - 1), where α = degree of dissociation and n = total particles on full dissociation.
i = 1 + 0.5 × (5 - 1) = 1 + 0.5 × 4 = 1 + 2 = 3

Step 4: Calculate osmotic pressure
Given,
M = 0.1 mol/L,
R = 0.0821 L atm K^-1 mol^-1 (convert units for atm),
T = 300 K,
i = 3.
So,
π = i × M × R × T = 3 × 0.1 × 0.0821 × 300 = 7.389 atm ≈ 7.48 atm

Step 5: Conclusion
The osmotic pressure of the 0.1 M potassium ferrocyanide solution at 300 K, 50% dissociated, is approximately 7.48 atm.