Two solid compounds X and Y dissociates at a certain temperature as follows
`X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)`
`Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)`
The total pressure of gases over a mixture of X and Y is `:`

Options

A. `4.5 atm`
B. `0.45 atm`
C. `0.6 atm`
D. None of these

Explanation

To solve the problem, we need to analyze the dissociation reactions of the solid compounds X and Y and then find the total pressure of the gases produced from these reactions.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ X(s) \rightleftharpoons A(g) + 2B(g) \quad K_{p1} = 9 \times 10^{-3} \, \text{atm}^3 \]
- The second reaction is:
\[ Y(s) \rightleftharpoons 2B(g) + C(g) \quad K_{p2} = 4.5 \times 10^{-3} \, \text{atm}^3 \]

2. **Combine the Reactions**:
- When we add the two reactions, we get:
\[ X(s) + Y(s) \rightleftharpoons A(g) + 4B(g) + C(g) \]

3. **Calculate the Overall Equilibrium Constant**:
- The overall equilibrium constant \( K_p \) for the combined reaction can be calculated as:
\[ K_p = K_{p1} \times K_{p2} \]
- Substituting the values:
\[ K_p = (9 \times 10^{-3}) \times (4.5 \times 10^{-3}) = 4.05 \times 10^{-5} \, \text{atm}^6 \]

4. **Define Partial Pressures**:
- Let the partial pressure of \( A \) be \( P_A = P \).
- The partial pressure of \( B \) will be \( P_B = 4P \) (since there are 4 moles of \( B \)).
- The partial pressure of \( C \) will be \( P_C = P \).

5. **Express \( K_p \) in Terms of Partial Pressures**:
- The expression for \( K_p \) in terms of partial pressures is:
\[ K_p = \frac{P_A \cdot (P_B)^4 \cdot P_C}{1} \]
- Substituting the values:
\[ K_p = P \cdot (4P)^4 \cdot P = P \cdot 256P^4 = 256P^5 \]

6. **Set Up the Equation**:
- Equating the two expressions for \( K_p \):
\[ 256P^5 = 4.05 \times 10^{-5} \]

7. **Solve for \( P \)**:
- Rearranging gives:
\[ P^5 = \frac{4.05 \times 10^{-5}}{256} \]
- Calculate \( P^5 \):
\[ P^5 = 1.58 \times 10^{-7} \]
- Taking the fifth root:
\[ P = (1.58 \times 10^{-7})^{1/5} \approx 0.0735 \, \text{atm} \]

8. **Calculate Total Pressure**:
- The total pressure \( P_{total} \) is the sum of the partial pressures:
\[ P_{total} = P_A + P_B + P_C = P + 4P + P = 6P \]
- Substituting \( P \):
\[ P_{total} = 6 \times 0.0735 = 0.441 \, \text{atm} \approx 0.45 \, \text{atm} \]

### Final Answer:
The total pressure of gases over a mixture of X and Y is approximately **0.45 atm**.