400 mL of oxygen are collected over water at N.T.P. `("Aqueous tension at "25^(@)C="0.0318 atm")`.calculate the volume of oxygen at 1 atm and 0 degree celcius.

Explanation

To calculate the volume of oxygen at 1 atm and 0 degrees Celsius, we will use the ideal gas law and the concept of gas laws. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Volume of oxygen collected over water (V1) = 400 mL
- Aqueous tension (water vapor pressure) at 25°C = 0.0318 atm
- Temperature (T1) = 25°C = 298 K (since T(K) = T(°C) + 273)
- Final pressure (P2) = 1 atm
- Final temperature (T2) = 0°C = 273 K

### Step 2: Calculate the pressure of the dry gas (P1)
The pressure of the dry gas (P1) can be calculated using the formula:
\[ P1 = P_{total} - P_{water} \]
Where:
- \( P_{total} \) = 1 atm (at NTP)
- \( P_{water} \) = 0.0318 atm

Thus,
\[ P1 = 1 \, \text{atm} - 0.0318 \, \text{atm} = 0.9682 \, \text{atm} \]

### Step 3: Use the combined gas law
The combined gas law relates the initial and final states of a gas:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]

### Step 4: Rearrange the equation to solve for V2
Rearranging gives:
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
- \( P1 = 0.9682 \, \text{atm} \)
- \( V1 = 400 \, \text{mL} \)
- \( T1 = 298 \, \text{K} \)
- \( T2 = 273 \, \text{K} \)
- \( P2 = 1 \, \text{atm} \)

So,
\[ V2 = \frac{0.9682 \, \text{atm} \cdot 400 \, \text{mL} \cdot 273 \, \text{K}}{1 \, \text{atm} \cdot 298 \, \text{K}} \]

### Step 6: Calculate V2
Calculating the above expression:
\[ V2 = \frac{0.9682 \cdot 400 \cdot 273}{298} \]

Calculating the numerator:
\[ 0.9682 \cdot 400 \cdot 273 = 105,272.4 \]

Now, divide by 298:
\[ V2 = \frac{105,272.4}{298} \approx 353.5 \, \text{mL} \]

Thus, the volume of oxygen at 1 atm and 0 degrees Celsius is approximately **353.5 mL**.