For equilibrium
`ZnSO_(4).7H_(2)O(s)hArrZnSO_(4).2H_(2)O(s)+5H_(2)O(g) K_(P)=56.25xx10^(-10) atm^(5)`
and vapour pressure of water is `22.8` torr at `298K.ZnSO_(4).7H_(2)O(s)` is efflorescent (lose water) when relative humidity is `[5sqrt56.25=2.23]`

Options

A. less than `70.620%`
B. less then `74.60%`
C. Above than `74.60%`
D. Above 70.60%`

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and calculate the relative humidity at which `ZnSO4.7H2O` loses water.

### Step 1: Write the equilibrium expression

The equilibrium reaction is given as:

\[ \text{ZnSO}_4 \cdot 7\text{H}_2\text{O}(s) \rightleftharpoons \text{ZnSO}_4 \cdot 2\text{H}_2\text{O}(s) + 5\text{H}_2\text{O}(g) \]

For this reaction, the equilibrium constant \( K_P \) is given by:

\[ K_P = \frac{P_{H_2O}^5}{1} \]

Where \( P_{H_2O} \) is the partial pressure of water vapor.

### Step 2: Substitute the value of \( K_P \)

We know that:

\[ K_P = 56.25 \times 10^{-10} \, \text{atm}^5 \]

Thus, we can set up the equation:

\[ P_{H_2O}^5 = 56.25 \times 10^{-10} \]

### Step 3: Calculate the partial pressure of water vapor

To find \( P_{H_2O} \), we take the fifth root of both sides:

\[ P_{H_2O} = \left( 56.25 \times 10^{-10} \right)^{1/5} \]

Calculating this gives:

\[ P_{H_2O} = \left( 56.25 \right)^{1/5} \times \left( 10^{-10} \right)^{1/5} \]

Calculating \( \left( 56.25 \right)^{1/5} \) yields approximately \( 2.23 \) and \( \left( 10^{-10} \right)^{1/5} = 10^{-2} \).

Thus,

\[ P_{H_2O} \approx 2.23 \times 10^{-2} \, \text{atm} \]

### Step 4: Convert vapor pressure of water to atm

The vapor pressure of water at 298 K is given as \( 22.8 \, \text{torr} \). To convert this to atm:

\[ P_{H_2O, \text{vapor}} = \frac{22.8 \, \text{torr}}{760 \, \text{torr/atm}} \approx 0.03 \, \text{atm} \]

### Step 5: Calculate the relative humidity

Relative humidity (RH) is defined as:

\[ RH = \frac{P_{H_2O}}{P_{H_2O, \text{vapor}}} \times 100\% \]

Substituting the values we calculated:

\[ RH = \frac{2.23 \times 10^{-2} \, \text{atm}}{0.03 \, \text{atm}} \times 100\% \]

Calculating this gives:

\[ RH \approx \frac{2.23}{3} \times 100\% \approx 74.33\% \]

### Step 6: Conclusion

The relative humidity at which \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \) loses water is approximately \( 74.33\% \).