Calculate the volume occupied by `10^(22)` molecules of a gas at 300 K and 760 mm pressure.

Explanation

To calculate the volume occupied by \(10^{22}\) molecules of a gas at 300 K and 760 mm pressure, we can follow these steps:

### Step 1: Convert the number of molecules to moles
We know that 1 mole of any substance contains \(6.022 \times 10^{23}\) molecules (Avogadro's number). Therefore, we can calculate the number of moles (\(n\)) from the number of molecules:

\[ n = \frac{N}{N_A} = \frac{10^{22}}{6.022 \times 10^{23}} \]

### Step 2: Calculate the number of moles
Now we perform the calculation:

\[ n = \frac{10^{22}}{6.022 \times 10^{23}} \approx 0.0166 \text{ moles} \]

### Step 3: Use the Ideal Gas Law
The Ideal Gas Law is given by the equation:

\[ PV = nRT \]

Where:
- \(P\) = pressure in atm
- \(V\) = volume in liters
- \(n\) = number of moles
- \(R\) = ideal gas constant = \(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- \(T\) = temperature in Kelvin

### Step 4: Convert pressure from mmHg to atm
Since the pressure is given as 760 mmHg, we convert it to atm:

\[ P = 1 \text{ atm} \quad (\text{since } 760 \text{ mmHg} = 1 \text{ atm}) \]

### Step 5: Substitute values into the Ideal Gas Law
Now we can substitute the values into the Ideal Gas Law:

\[ 1 \text{ atm} \cdot V = 0.0166 \text{ moles} \cdot 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 300 \text{ K} \]

### Step 6: Calculate the volume \(V\)
Now we calculate \(V\):

\[ V = \frac{0.0166 \cdot 0.0821 \cdot 300}{1} \]

Calculating the right side:

\[ V \approx 0.409 \text{ L} \]

### Step 7: Convert volume to cm³
To convert liters to cubic centimeters (cm³), we use the conversion factor \(1 \text{ L} = 1000 \text{ cm}^3\):

\[ V \approx 0.409 \text{ L} \times 1000 \text{ cm}^3/\text{L} = 409 \text{ cm}^3 \]

### Final Answer
The volume occupied by \(10^{22}\) molecules of the gas at 300 K and 760 mm pressure is approximately:

\[ \text{Volume} \approx 0.409 \text{ L} \quad \text{or} \quad 409 \text{ cm}^3 \]