Decinormal solution of NaCl developed an osmotic pressure of 4.6 atmosphere at 300 K. Calcualte its degree of dissociatoin (R = 0.082`" L atm K"^(-1)"mol"^(-1)`)

Answer

To solve the problem of calculating the degree of dissociation of a decinormal solution of NaCl that developed an osmotic pressure of 4.6 atm at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Osmotic pressure (π) = 4.6 atm
- Temperature (T) = 300 K
- Gas constant (R) = 0.0821 L atm K⁻¹ mol⁻¹
- Normality of NaCl solution = 0.1 N (decinormal)

2. **Convert Normality to Molarity:**
- Molarity (C) = Normality / n-factor
- For NaCl, the n-factor = 1 (since it dissociates into Na⁺ and Cl⁻).
- Therefore, Molarity (C) = 0.1 N / 1 = 0.1 M.

3. **Use the Osmotic Pressure Formula:**
- The formula for osmotic pressure is given by:
\[ \pi = C \cdot R \cdot T \cdot i \]
- Where:
- π = osmotic pressure
- C = molarity of the solution
- R = gas constant
- T = temperature in Kelvin
- i = van 't Hoff factor (which accounts for the degree of dissociation).

4. **Rearranging the Formula to Find i:**
- Rearranging the formula gives:
\[ i = \frac{\pi}{C \cdot R \cdot T} \]

5. **Substituting the Known Values:**
- Substitute π = 4.6 atm, C = 0.1 M, R = 0.0821 L atm K⁻¹ mol⁻¹, and T = 300 K into the equation:
\[ i = \frac{4.6}{0.1 \cdot 0.0821 \cdot 300} \]

6. **Calculating i:**
- Calculate the denominator:
\[ 0.1 \cdot 0.0821 \cdot 300 = 2.463 \]
- Now calculate i:
\[ i = \frac{4.6}{2.463} \approx 1.86 \]

7. **Finding the Degree of Dissociation (α):**
- The van 't Hoff factor (i) is related to the degree of dissociation (α) by the formula:
\[ i = 1 + \alpha(n - 1) \]
- For NaCl, n = 2 (since it dissociates into Na⁺ and Cl⁻).
- Substitute i = 1.86 and n = 2 into the equation:
\[ 1.86 = 1 + \alpha(2 - 1) \]
- Simplifying gives:
\[ 1.86 = 1 + \alpha \]
- Therefore:
\[ \alpha = 1.86 - 1 = 0.86 \]

8. **Convert α to Percentage:**
- To express the degree of dissociation as a percentage:
\[ \alpha \times 100 = 0.86 \times 100 = 86\% \]

### Final Answer:
The degree of dissociation (α) of the decinormal solution of NaCl is **86%**.