`CuSO_(4) .5H_(2)O_((g)) hArr CuSO_(4), 3H_(2)O_((s)) + 2H_(2) O_((g)) , K_(p) = 4 xx 10^(-4) atm^(2)`. If the vapour pressure of wateris 38 torr then percentage of relative humidity is: (Assume all data at constant temperature)

Options

A. 4
B. 10
C. 40
D. 1

Answer

To solve the problem, we will follow these steps:

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

The equilibrium constant \( K_p \) for this reaction is given as:
\[ K_p = 4 \times 10^{-4} \, \text{atm}^2 \]

### Step 2: Relate Kp to Partial Pressure
For the reaction, the expression for \( K_p \) is:
\[ K_p = \frac{(P_{H_2O})^2}{1} \]
where \( P_{H_2O} \) is the partial pressure of water vapor at equilibrium.

### Step 3: Calculate the Partial Pressure of Water
From the \( K_p \) value, we can find the partial pressure of water:
\[ P_{H_2O} = \sqrt{K_p} = \sqrt{4 \times 10^{-4}} \]
\[ P_{H_2O} = 2 \times 10^{-2} \, \text{atm} = 0.02 \, \text{atm} \]

### Step 4: Convert Vapor Pressure from Torr to atm
The vapor pressure of water is given as 38 torr. We need to convert this to atm:
\[ 1 \, \text{atm} = 760 \, \text{torr} \]
Thus,
\[ P_{vapor} = \frac{38 \, \text{torr}}{760 \, \text{torr/atm}} = 0.05 \, \text{atm} \]

### Step 5: Calculate Relative Humidity
Relative humidity (RH) is calculated using the formula:
\[ RH = \left( \frac{P_{H_2O}}{P_{vapor}} \right) \times 100 \]
Substituting the values we found:
\[ RH = \left( \frac{0.02 \, \text{atm}}{0.05 \, \text{atm}} \right) \times 100 \]
\[ RH = 0.4 \times 100 = 40\% \]

### Conclusion
The percentage of relative humidity is:
\[ \text{Relative Humidity} = 40\% \]