Decimolar solution of potassium ferricyanide, `K_3[Fe(CN)_6 ]` has osmotic pressure of 3.94 atm at `27^@C`. Hence percent ionisation of the solute is -

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer

To solve the problem of finding the percent ionization of potassium ferricyanide, we will follow these steps:

### Step 1: Understand the given data
We have a decimolar (0.1 M) solution of potassium ferricyanide, \( K_3[Fe(CN)_6] \), with an osmotic pressure (\( \pi \)) of 3.94 atm at a temperature of 27°C.

### Step 2: Convert temperature to Kelvin
To use the ideal gas law in the osmotic pressure formula, we need to convert the temperature from Celsius to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 3: Use the osmotic pressure formula
The formula for osmotic pressure is given by:
\[ \pi = i \cdot C \cdot R \cdot T \]
Where:
- \( \pi \) = osmotic pressure (3.94 atm)
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = concentration (0.1 M)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (300 K)

### Step 4: Rearrange the formula to find \( i \)
Rearranging the formula gives:
\[ i = \frac{\pi}{C \cdot R \cdot T} \]
Substituting the values:
\[ i = \frac{3.94}{0.1 \cdot 0.0821 \cdot 300} \]

### Step 5: Calculate \( i \)
Calculating the denominator:
\[ 0.1 \cdot 0.0821 \cdot 300 = 2.463 \]
Now substituting back:
\[ i = \frac{3.94}{2.463} \approx 1.60 \]

### Step 6: Determine the number of ions formed (n)
For potassium ferricyanide, the dissociation is:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]
This results in a total of 4 ions (3 potassium ions and 1 ferricyanide ion).

Thus, \( n = 4 \).

### Step 7: Calculate the degree of ionization (\( \alpha \))
The relationship between \( i \), \( n \), and \( \alpha \) is given by:
\[ i = 1 + \alpha(n - 1) \]
Substituting the known values:
\[ 1.60 = 1 + \alpha(4 - 1) \]
\[ 1.60 = 1 + 3\alpha \]
\[ 0.60 = 3\alpha \]
\[ \alpha = \frac{0.60}{3} = 0.20 \]

### Step 8: Calculate percent ionization
Percent ionization is given by:
\[ \text{Percent Ionization} = \alpha \times 100\% \]
Substituting the value of \( \alpha \):
\[ \text{Percent Ionization} = 0.20 \times 100\% = 20\% \]

### Final Answer
The percent ionization of the solute is **20%**.

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