In the presence of excess of anhydrous ( in torr) of water taken up is governed by `K_(p)=10^(12)atm^(-4)` for the following reaction at `273K`
`SrCl_(2).2H_(2)O(s)+4H_(2)O(g)hArrSrCl_(2).6H_(2)O(s)`
What is equilibrium vapour pressure ( in torr) of water in a closedvessel that contains `SrCl_(2).2H_(2)O(s)` ?

Options

A. `0.001`torr`
B. `10^(3)` torr`
C. `0.76 `torr
D. `1.31 `torr`

Answer

To solve the problem, we need to find the equilibrium vapor pressure of water in a closed vessel containing SrCl₂·2H₂O(s) at 273 K, given that the equilibrium constant \( K_p = 10^{12} \, \text{atm}^{-4} \) for the reaction:

\[ \text{SrCl}_2·2\text{H}_2\text{O}(s) + 4\text{H}_2\text{O}(g) \rightleftharpoons \text{SrCl}_2·6\text{H}_2\text{O}(s) \]

### Step-by-Step Solution:

1. **Write the expression for \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by the formula:
\[ K_p = \frac{P_{\text{products}}}{P_{\text{reactants}}} \]
For the given reaction, the expression simplifies to:
\[ K_p = \frac{1}{P_{\text{H}_2\text{O}}^4} \]
where \( P_{\text{H}_2\text{O}} \) is the partial pressure of water vapor.

2. **Substitute the value of \( K_p \)**:
We know that \( K_p = 10^{12} \, \text{atm}^{-4} \). Therefore, we can write:
\[ 10^{12} = \frac{1}{P_{\text{H}_2\text{O}}^4} \]

3. **Rearrange to find \( P_{\text{H}_2\text{O}} \)**:
Rearranging the equation gives:
\[ P_{\text{H}_2\text{O}}^4 = \frac{1}{10^{12}} \]
Taking the fourth root of both sides:
\[ P_{\text{H}_2\text{O}} = \left(10^{-12}\right)^{1/4} = 10^{-3} \, \text{atm} \]

4. **Convert \( P_{\text{H}_2\text{O}} \) to torr**:
To convert the pressure from atm to torr, we use the conversion factor \( 1 \, \text{atm} = 760 \, \text{torr} \):
\[ P_{\text{H}_2\text{O}} = 10^{-3} \, \text{atm} \times 760 \, \text{torr/atm} = 0.76 \, \text{torr} \]

5. **Final Answer**:
The equilibrium vapor pressure of water in the closed vessel is:
\[ \boxed{0.76 \, \text{torr}} \]