The osmotic pressure of a decinormal solution of `BaCl_(2)` at `27^(@)C` showing 80% degree of ionisation will be :

Options

A. 3.20 atm
B. 4.20 atm
C. 5. 20 atm
D. 6.20 atm

Answer

To find the osmotic pressure of a decinormal solution of BaCl₂ at 27°C with an 80% degree of ionization, we can follow these steps:

### Step 1: Identify Given Values
- Normality (N) = 0.1 N (decinormal solution)
- Degree of ionization (α) = 80% = 0.8
- Temperature (T) = 27°C = 300 K (after converting to Kelvin)

### Step 2: Calculate the Van 't Hoff Factor (I)
The Van 't Hoff factor (I) can be calculated using the formula:
\[ I = 1 + (N - 1) \cdot \alpha \]
Where:
- N = number of ions produced from the dissociation of the solute.

For BaCl₂:
- Barium (Ba²⁺) = 1 ion
- Chloride (Cl⁻) = 2 ions

Thus, N = 1 + 2 = 3.

Now substituting the values:
\[ I = 1 + (3 - 1) \cdot 0.8 \]
\[ I = 1 + 2 \cdot 0.8 \]
\[ I = 1 + 1.6 \]
\[ I = 2.6 \]

### Step 3: Convert Normality to Molarity
Normality (N) is related to molarity (C) by the equation:
\[ \text{Normality} = \text{Molarity} \times \text{N factor} \]

For BaCl₂, the N factor is 2 (as it provides 2 moles of Cl⁻ ions). Therefore:
\[ C = \frac{N}{\text{N factor}} \]
\[ C = \frac{0.1}{2} \]
\[ C = 0.05 \, \text{M} \]

### Step 4: Use the Osmotic Pressure Formula
The formula for osmotic pressure (π) is given by:
\[ \pi = I \cdot C \cdot R \cdot T \]
Where:
- R = 0.0821 L·atm/(K·mol) (universal gas constant)
- T = 300 K (temperature in Kelvin)

Now substituting the values:
\[ \pi = 2.6 \cdot 0.05 \cdot 0.0821 \cdot 300 \]

### Step 5: Calculate Osmotic Pressure
Calculating the above expression:
\[ \pi = 2.6 \cdot 0.05 \cdot 0.0821 \cdot 300 \]
\[ \pi = 2.6 \cdot 0.05 \cdot 24.63 \]
\[ \pi = 2.6 \cdot 1.2315 \]
\[ \pi = 3.20 \, \text{atm} \]

### Final Answer
The osmotic pressure of the decinormal solution of BaCl₂ at 27°C showing 80% degree of ionization is **3.20 atm**.

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