If osmotic presssure of 1M aqueous solution of `H_(2)SO_(4)` at 500 K is 90.2 atm. Calculate `K_(a2)` of `H_(2)SO_(4)`. Give your answer after multiplying 1000 with `K_(a_(2))`. (Assumme ideal solution) [Given : `K_(a_(1))` of `H_(2)SO_(4)=oo`,R=0.082 L atm`//`mol-K]

Explanation

To solve the problem, we need to calculate the second dissociation constant \( K_{a2} \) of sulfuric acid \( H_2SO_4 \) based on the given osmotic pressure of a 1M solution at 500 K.

### Step-by-Step Solution:

1. **Write the Dissociation Reaction:**
The dissociation of sulfuric acid can be represented as:
\[ H_2SO_4 \rightleftharpoons H^+ + HSO_4^- \]
\[ HSO_4^- \rightleftharpoons H^+ + SO_4^{2-} \]

2. **Define Initial Concentrations:**
For the initial concentration of \( H_2SO_4 \), we have:
- At \( t = 0 \):
- \( [H_2SO_4] = 1 \, \text{M} \)
- \( [H^+] = 0 \)
- \( [HSO_4^-] = 0 \)
- \( [SO_4^{2-}] = 0 \)

3. **Set Up the Equilibrium Expression:**
After the first dissociation, we assume \( x \) is the amount that dissociates:
- At equilibrium:
- \( [H_2SO_4] = 1 - x \)
- \( [H^+] = x \)
- \( [HSO_4^-] = x \)

4. **Calculate Total Concentration:**
The total concentration of ions at equilibrium is:
\[ C = [H^+] + [HSO_4^-] + [SO_4^{2-}] = x + x + (1 - x) = 1 + x \]

5. **Use the Osmotic Pressure Formula:**
The osmotic pressure \( \pi \) is given by the formula:
\[ \pi = C \cdot R \cdot T \]
Substituting the known values:
\[ 90.2 \, \text{atm} = (1 + x) \cdot 0.082 \, \text{L atm} / \text{mol K} \cdot 500 \, \text{K} \]

6. **Solve for \( x \):**
Rearranging the equation:
\[ 90.2 = (1 + x) \cdot 41 \]
\[ 1 + x = \frac{90.2}{41} \approx 2.2 \]
\[ x \approx 1.2 \]

7. **Determine Concentrations of Ions:**
Now we can find the concentrations:
- \( [H^+] = 1 + x = 1 + 1.2 = 2.2 \, \text{M} \)
- \( [HSO_4^-] = x = 1.2 \, \text{M} \)
- \( [SO_4^{2-}] = 1 - x = 1 - 1.2 = -0.2 \, \text{M} \) (not possible, so we need to adjust our calculations)

8. **Calculate \( K_{a2} \):**
The expression for \( K_{a2} \) is:
\[ K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} \]
Substituting the values:
\[ K_{a2} = \frac{(1.2)(0.2)}{0.8} = \frac{0.24}{0.8} = 0.3 \]

9. **Final Calculation:**
Finally, we multiply \( K_{a2} \) by 1000:
\[ K_{a2} \times 1000 = 0.3 \times 1000 = 300 \]

### Final Answer:
\[ \text{The value of } K_{a2} \text{ multiplied by 1000 is } 300. \]