For the equilibrium `CuSO_(4)xx5H_(2)O(s)hArrCuSO_(4)xx3H_(2)O(s) + 2H_(2)O(g)`
`K_(p) = 2.25xx10^(-4)atm^(2)` and vapour pressure of water is `22.8` torr at 298 K. `CuSO_(4)` . `5H_(2)O(s)` is efflorescent (i.e., losses water) when relative humidity is :

Options

A. less than `33.3%`
B. less than `50%`
C. less than `66.6%`
D. above `66.6%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction given is:
\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4 \cdot 3\text{H}_2\text{O}(s) + 2\text{H}_2\text{O}(g) \]

### Step 2: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_p = P_{\text{H}_2\text{O}}^2 \]
where \( P_{\text{H}_2\text{O}} \) is the partial pressure of water vapor.

### Step 3: Substitute the given \( K_p \) value
We are given that:
\[ K_p = 2.25 \times 10^{-4} \, \text{atm}^2 \]

### Step 4: Calculate the partial pressure of \( \text{H}_2\text{O} \)
To find the partial pressure of water vapor, we take the square root of \( K_p \):
\[ P_{\text{H}_2\text{O}} = \sqrt{2.25 \times 10^{-4}} \]
Calculating this gives:
\[ P_{\text{H}_2\text{O}} = 1.5 \times 10^{-2} \, \text{atm} \]

### Step 5: Convert the vapor pressure of water from torr to atm
The vapor pressure of water at 298 K is given as 22.8 torr. To convert this to atm, we use the conversion factor \( 1 \, \text{atm} = 760 \, \text{torr} \):
\[ P_{\text{H}_2\text{O, vapor}} = \frac{22.8 \, \text{torr}}{760 \, \text{torr/atm}} = 0.03 \, \text{atm} \]

### Step 6: Calculate the relative humidity
Relative humidity (RH) is calculated using the formula:
\[ \text{RH} = \frac{P_{\text{H}_2\text{O}}}{P_{\text{H}_2\text{O, vapor}}} \times 100 \]
Substituting the values we found:
\[ \text{RH} = \frac{1.5 \times 10^{-2} \, \text{atm}}{0.03 \, \text{atm}} \times 100 \]
Calculating this gives:
\[ \text{RH} = 50\% \]

### Step 7: Determine the condition for efflorescence
Since CuSO4·5H2O is efflorescent, it loses water when the relative humidity is less than a certain value. From our calculation, we find that it will lose water when the relative humidity is less than 50%.

### Final Answer
The relative humidity at which CuSO4·5H2O is efflorescent is less than 50%.

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