A definite amount of solid `NH_(4)HS` is placed in a flask already containing ammonia gas at a certain temperature and 0.1 atm pressure. `NH_(4)HS` decompses to give ` NH_(3)` and `H_(2)S` and at equilibrium total pressure in flask is 1.1 atm. If the equilibrium constant `K_(P)` for the reaction `NH_(4)HS(s) iff NH_(3)(g)+H_(2)S(g)` is represented as `zxx10^(-1)` then find the value of z.

Answer

To solve the problem, we need to analyze the decomposition of solid ammonium hydrosulfide (NH₄HS) into ammonia (NH₃) and hydrogen sulfide (H₂S) at equilibrium. We are given the initial pressure of ammonia and the total pressure at equilibrium, and we need to find the equilibrium constant \( K_p \) in the form \( z \times 10^{-1} \).

### Step-by-step Solution:

1. **Write the decomposition reaction:**
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

2. **Identify initial conditions:**
- Initial pressure of NH₃ = 0.1 atm
- Initial pressure of H₂S = 0 atm (since it is produced from the decomposition)
- Since NH₄HS is a solid, it does not contribute to the pressure.

3. **Define changes at equilibrium:**
- Let \( x \) be the change in pressure due to the decomposition of NH₄HS.
- At equilibrium:
- Pressure of NH₃ = \( 0.1 + x \)
- Pressure of H₂S = \( x \)

4. **Total pressure at equilibrium:**
- Given total pressure = 1.1 atm
\[ (0.1 + x) + x = 1.1 \]
\[ 0.1 + 2x = 1.1 \]
\[ 2x = 1.1 - 0.1 = 1.0 \]
\[ x = 0.5 \]

5. **Calculate equilibrium pressures:**
- Pressure of NH₃ at equilibrium = \( 0.1 + 0.5 = 0.6 \) atm
- Pressure of H₂S at equilibrium = \( 0.5 \) atm

6. **Write the expression for the equilibrium constant \( K_p \):**
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{1} = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]
\[ K_p = (0.6)(0.5) = 0.3 \]

7. **Express \( K_p \) in the required form:**
\[ K_p = 3 \times 10^{-1} \]
Here, \( z = 3 \).

### Final Answer:
The value of \( z \) is **3**.