A decimolar solution of potassium ferrocyanide is `50%` dissociated at 300K. Calculate the osmotic pressure of the solution. `(R="8.314 J K"^(-1)"mol"^(-1)).`

Explanation

To calculate the osmotic pressure of a decimolar solution of potassium ferrocyanide that is 50% dissociated at 300 K, we can follow these steps:

### Step 1: Understand the given data
- Concentration (C) = 0.1 M (decimolar solution)
- Degree of dissociation (α) = 0.5 (50% dissociation)
- Temperature (T) = 300 K
- Gas constant (R) = 8.314 J K⁻¹ mol⁻¹ (we will convert it to L atm K⁻¹ mol⁻¹ for our calculations)

### Step 2: Determine the van 't Hoff factor (i)
Potassium ferrocyanide (K₄[Fe(CN)₆]) dissociates in solution as follows:
\[ K₄[Fe(CN)₆] \rightarrow 4K^+ + [Fe(CN)₆]^{4-} \]

From the dissociation, we see that:
- 4 K⁺ ions and 1 [Fe(CN)₆]⁴⁻ ion are produced.
- Total ions produced = 4 + 1 = 5 ions.

The van 't Hoff factor (i) is given by the formula:
\[ i = 1 + \alpha(n - 1) \]
where n is the number of particles produced from one formula unit.

Here, n = 5 (total ions), and α = 0.5. Therefore:
\[ i = 1 + 0.5(5 - 1) \]
\[ i = 1 + 0.5 \times 4 \]
\[ i = 1 + 2 = 3 \]

### Step 3: Use the osmotic pressure formula
The formula for osmotic pressure (π) is:
\[ \pi = iCRT \]

Substituting the values we have:
- C = 0.1 M
- R = 0.0821 L atm K⁻¹ mol⁻¹ (converted from J K⁻¹ mol⁻¹)
- T = 300 K
- i = 3

### Step 4: Calculate the osmotic pressure
Now we can plug in the values:
\[ \pi = 3 \times 0.1 \, \text{mol/L} \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} \]

Calculating this:
\[ \pi = 3 \times 0.1 \times 0.0821 \times 300 \]
\[ \pi = 3 \times 0.1 \times 24.63 \]
\[ \pi = 3 \times 2.463 \]
\[ \pi = 7.389 \, \text{atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **7.389 atm**.

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