A certain quantity of a gas occupies 300 mL when collected over water at `25^@C` and 745 mm pressure. It occupied 182.6 mL in dry state at S.T.P. Find the vapour pressure of water at `25^@C`.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between the pressures and volumes of the gas at different conditions. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Volume of gas collected over water (V1) = 300 mL = 0.3 L
- Pressure of gas collected over water (P1) = 745 mmHg
- Volume of dry gas at STP (V2) = 182.6 mL = 0.1826 L
- Standard pressure at STP (P2) = 760 mmHg
- Temperature when gas was collected (T1) = 25°C = 298 K
- Temperature at STP (T2) = 0°C = 273 K

### Step 2: Set Up the Equation
Since the gas is collected over water, the pressure of the dry gas (P_dry) can be expressed as:
\[ P_{dry} = P_1 - P_{water} \]
where \( P_{water} \) is the vapor pressure of water at 25°C.

### Step 3: Use the Ideal Gas Law
According to the ideal gas law, we can set up the equation:
\[ \frac{P_{dry} \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ \frac{(745 - P_{water}) \cdot 0.3}{298} = \frac{760 \cdot 0.1826}{273} \]

### Step 5: Calculate the Right Side
Calculating the right side:
\[ \frac{760 \cdot 0.1826}{273} = \frac{138.776}{273} \approx 0.508 \]

### Step 6: Rearranging the Equation
Now, we can rearrange the equation to find \( P_{water} \):
\[ (745 - P_{water}) \cdot 0.3 = 0.508 \cdot 298 \]
\[ 745 - P_{water} = \frac{0.508 \cdot 298}{0.3} \]

### Step 7: Calculate the Left Side
Calculating the left side:
\[ 0.508 \cdot 298 \approx 151.784 \]
\[ 745 - P_{water} = \frac{151.784}{0.3} \approx 505.947 \]

### Step 8: Solve for \( P_{water} \)
Now, solving for \( P_{water} \):
\[ P_{water} = 745 - 505.947 \]
\[ P_{water} \approx 239.053 \text{ mmHg} \]

### Final Answer
The vapor pressure of water at 25°C is approximately **239.05 mmHg**.

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