`CuSO_(4).5H_(2)O(s)hArrCuSO_(4). 3H_(2)O(s)+2H_(2)O(g), K_(p)=4xx10^(-4)atm^(2)` If the vapour pressure of water is 38 toor then percentage of relatative humidity is :(Assume all data at constant temperture)

Options

A. 4
B. 10
C. 40
D. none of these

Answer

To solve the problem, we need to determine the percentage of relative humidity based on the given equilibrium reaction and the vapor pressure of water. Let's break down the steps:

### Step-by-Step Solution:

1. **Write the Equilibrium Expression**:
The equilibrium reaction is:
\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4 \cdot 3\text{H}_2\text{O}(s) + 2\text{H}_2\text{O}(g) \]
The equilibrium constant \( K_p \) is given as:
\[ K_p = 4 \times 10^{-4} \text{ atm}^2 \]
Since solids do not appear in the equilibrium expression, we only consider the gaseous component:
\[ K_p = \frac{(P_{\text{H}_2\text{O}})^2}{1} = (P_{\text{H}_2\text{O}})^2 \]

2. **Calculate the Partial Pressure of Water**:
To find the partial pressure of water, we take the square root of \( K_p \):
\[ P_{\text{H}_2\text{O}} = \sqrt{K_p} = \sqrt{4 \times 10^{-4}} = 2 \times 10^{-2} \text{ atm} \]

3. **Convert Partial Pressure to mmHg**:
We know that 1 atm = 760 mmHg, so we convert the partial pressure of water to mmHg:
\[ P_{\text{H}_2\text{O}} = 2 \times 10^{-2} \text{ atm} \times 760 \text{ mmHg/atm} = 15.2 \text{ mmHg} \]

4. **Calculate Relative Humidity**:
The formula for relative humidity (RH) is given by:
\[ \text{RH} = \left( \frac{P_{\text{H}_2\text{O}}}{P_{\text{H}_2\text{O, saturation}}} \right) \times 100 \]
Given that the vapor pressure of water is 38 torr, we can substitute:
\[ \text{RH} = \left( \frac{15.2 \text{ mmHg}}{38 \text{ mmHg}} \right) \times 100 \]

5. **Perform the Calculation**:
\[ \text{RH} = \left( \frac{15.2}{38} \right) \times 100 \approx 40\% \]

### Final Answer:
The percentage of relative humidity is **40%**.

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