`CuSO_(4),5H_(2)O(s)hArrCuSO_(4)(s)+5H_(2)O(g)K_(P)=10^(-10) "moles of" CuSO_(4).5H_(2)O(s)` is taken in a `2.5L` container at `27^(@)C` then at equilibrium [Take: `R=(1)/(12)` litre atm `mol^(-1)K^(-1)`]

Options

A. Moles of `CuSO_(4).5H_(2)O` left in the container is `9xx10^(-3)`
B. Moles of `CuSO_(4).5H_(2)O` left in the container is `9.8xx10^(-3)`
C. Moles of `CuSO_(4)` Left in the container is `10^(-3)`
D. Moles of `CuSO_(4)` left in the container is `2xx10^(-4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Reaction
The reaction given is:
\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g) \]
Here, we have solid CuSO4·5H2O on the left, which does not affect the equilibrium constant, and gaseous water on the right.

### Step 2: Identify Given Data
- \( K_p = 10^{-10} \)
- Volume of container = 2.5 L
- Temperature = 27°C = 300 K (after converting to Kelvin)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

### Step 3: Calculate \( \Delta n_g \)
The change in the number of moles of gas (\( \Delta n_g \)) is calculated as follows:
- Moles of gaseous products = 5 (from 5H2O)
- Moles of gaseous reactants = 0 (since there are no gaseous reactants)
Thus,
\[ \Delta n_g = 5 - 0 = 5 \]

### Step 4: Relate \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c \cdot R^T \cdot \Delta n_g \]
We can rearrange this to find \( K_c \):
\[ K_c = \frac{K_p}{R^T \cdot \Delta n_g} \]

### Step 5: Calculate \( K_c \)
Substituting the values:
\[ K_c = \frac{10^{-10}}{(0.0821)(300)(5)} \]
Calculating the denominator:
\[ R \cdot T \cdot \Delta n_g = 0.0821 \cdot 300 \cdot 5 = 123.15 \]
Now, substituting back:
\[ K_c = \frac{10^{-10}}{123.15} \approx 8.12 \times 10^{-13} \]

### Step 6: Write the Expression for \( K_c \)
The equilibrium expression for the reaction is:
\[ K_c = \frac{[\text{H}_2\text{O}]^5}{1} \]
Thus,
\[ K_c = [\text{H}_2\text{O}]^5 \]

### Step 7: Calculate the Concentration of \( \text{H}_2\text{O} \)
From the expression:
\[ [\text{H}_2\text{O}]^5 = 8.12 \times 10^{-13} \]
Taking the fifth root:
\[ [\text{H}_2\text{O}] = (8.12 \times 10^{-13})^{1/5} \approx 1.65 \times 10^{-3} \, \text{mol/L} \]

### Step 8: Calculate Moles of \( \text{H}_2\text{O} \)
Using the concentration to find the total moles in the 2.5 L container:
\[ \text{Moles of } \text{H}_2\text{O} = [\text{H}_2\text{O}] \times \text{Volume} = 1.65 \times 10^{-3} \, \text{mol/L} \times 2.5 \, \text{L} = 4.125 \times 10^{-3} \, \text{mol} \]

### Step 9: Relate Moles of \( \text{H}_2\text{O} \) to \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)
From the stoichiometry of the reaction, 1 mole of CuSO4·5H2O produces 5 moles of H2O. Therefore, the moles of CuSO4·5H2O that reacted to produce this amount of H2O is:
\[ \text{Moles of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = \frac{4.125 \times 10^{-3}}{5} = 8.25 \times 10^{-4} \]

### Step 10: Calculate Moles of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) Remaining
Initially, we had \( 10^{-2} \) moles of CuSO4·5H2O (since it is not specified, we assume it is 0.01 moles). Therefore, the moles remaining at equilibrium are:
\[ \text{Moles remaining} = 10^{-2} - 8.25 \times 10^{-4} = 9.175 \times 10^{-3} \]

### Final Answer
The moles of CuSO4·5H2O left in the container at equilibrium is approximately \( 9.175 \times 10^{-3} \) moles.