A 1% (mass/vol) KCl solution is ionised to the extent of 80%. The osmotic pressure at `27^(@)`C of the solution will be :

Options

A. 6.95 atm
B. 5.94 atm
C. 2.71 atm
D. 3.30 atm

Answer

To solve the problem of calculating the osmotic pressure of a 1% (mass/vol) KCl solution that is ionized to the extent of 80%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- We have a 1% (mass/volume) KCl solution.
- The degree of ionization (α) is 80% or 0.8.
- The temperature is given as 27°C.

2. **Convert the Percentage to Weight:**
- A 1% (mass/volume) solution means there is 1 gram of KCl in 100 mL of solution.

3. **Calculate the Molar Mass of KCl:**
- The molar mass of KCl (Potassium Chloride) is approximately 74.5 g/mol.

4. **Calculate the Number of Moles of KCl:**
- Number of moles (n) = weight (g) / molar mass (g/mol).
- Here, n = 1 g / 74.5 g/mol = 0.0134 moles.

5. **Calculate the Concentration (C) in Moles per Liter:**
- Since we have 100 mL of solution, we convert it to liters: 100 mL = 0.1 L.
- Concentration (C) = moles / volume (L) = 0.0134 moles / 0.1 L = 0.134 M.

6. **Determine the Van't Hoff Factor (i):**
- KCl dissociates into K⁺ and Cl⁻. Therefore, the total number of ions produced = 2.
- However, since the solution is ionized to 80%, we calculate the effective number of particles:
- i = 1 + α = 1 + 0.8 = 1.8.

7. **Convert Temperature to Kelvin:**
- Temperature in Kelvin (T) = 27°C + 273 = 300 K.

8. **Use the Osmotic Pressure Formula:**
- The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \cdot i \]
- Where R (the gas constant) = 0.0821 L·atm/(K·mol).

9. **Substitute the Values into the Formula:**
- π = 0.134 M × 0.0821 L·atm/(K·mol) × 300 K × 1.8.
- π = 0.134 × 0.0821 × 300 × 1.8.

10. **Calculate the Osmotic Pressure:**
- π = 0.134 × 0.0821 × 300 × 1.8 = 5.94 atm.

### Final Answer:
The osmotic pressure of the solution is **5.94 atm**.

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