`CuSO_(4).5H_(2)O(s)hArrCuSO_(4).3H_(2)O(s)+2H_(2)O(s) K_(P)=0.4xx10^(-3) atm^(2)`
Which of following sttement are correct:

Options

A. `DeltaG^(@)=-RTlnP_(H_(2)o` where `P_(H_(2)o="Partial pressure of" H_(2)O` at equilibrium.
B. At vapour pressure of `H_(2)O=15.2` torr relative humidity of `CuSO_(4).5H_(2)O "is" 100%`
C. In presence of aqueous tension of `24` torr, `CuSO_(4).5H_(2)O` can not loss molisture.
D. In presence of dry atmosphere in open container `CuSO_(4).5H_(2)O` will completely convert into `CuSO_(4).3H_(2)O`

Explanation

To analyze the given question regarding the equilibrium of the reaction \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4 \cdot 3\text{H}_2\text{O}(s) + 2\text{H}_2\text{O}(g) \) with \( K_p = 0.4 \times 10^{-3} \, \text{atm}^2 \), we need to evaluate the correctness of the provided statements.

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constant**:
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products:
\[ K_p = \frac{(P_{H_2O})^2}{1} = P_{H_2O}^2 \]
Given \( K_p = 0.4 \times 10^{-3} \, \text{atm}^2 \), we can find the equilibrium partial pressure of water vapor, \( P_{H_2O} \):
\[ P_{H_2O} = \sqrt{K_p} = \sqrt{0.4 \times 10^{-3}} \approx 0.020 \, \text{atm} \]

2. **Evaluating Statement 1**:
The first statement claims:
\[ \Delta G^0 = -RT \ln P_{H_2O} \]
This statement is generally correct as it relates the standard Gibbs free energy change to the partial pressure of water at equilibrium. However, it should be noted that this is valid under certain conditions. Therefore, we will consider this statement as correct.

3. **Evaluating Statement 2**:
The second statement states that at a vapor pressure of \( 15.2 \, \text{torr} \), the relative humidity of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is 100%. We need to convert the equilibrium pressure we calculated into torr:
\[ 0.020 \, \text{atm} \times 760 \, \text{torr/atm} \approx 15.2 \, \text{torr} \]
Since the equilibrium pressure of water vapor at 100% relative humidity equals the vapor pressure, this statement is also correct.

4. **Evaluating Statement 3**:
The third statement claims that in the presence of a vapor pressure of \( 24 \, \text{torr} \), \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) cannot lose moisture. Since \( 24 \, \text{torr} \) is greater than \( 15.2 \, \text{torr} \), the reaction will shift backward, indicating that moisture cannot be lost. Therefore, this statement is correct.

5. **Evaluating Statement 4**:
The fourth statement claims that in a dry atmosphere in an open container, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) will completely convert into \( \text{CuSO}_4 \cdot 3\text{H}_2\text{O} \). In a dry atmosphere, the partial pressure of water vapor is significantly lower than the equilibrium pressure, leading to a forward shift in the reaction. Thus, this statement is correct.

### Conclusion:
The correct statements are:
- Statement 1: Correct
- Statement 2: Correct
- Statement 3: Correct
- Statement 4: Correct