Calculate the volume occupied by `4.045 xx 10^(2)` molecules of oxygen at `27^(@)C` and having a pressure of 700 torr.

Explanation

To calculate the volume occupied by \(4.045 \times 10^{23}\) molecules of oxygen at \(27^\circ C\) and a pressure of 700 torr, we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Given \(T = 27^\circ C\):
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Convert the pressure from torr to atm
To convert the pressure from torr to atm, we use the conversion factor \(1 \, atm = 760 \, torr\):
\[ P(atm) = \frac{P(torr)}{760} \]
Given \(P = 700 \, torr\):
\[ P = \frac{700}{760} \approx 0.9211 \, atm \]

### Step 3: Calculate the number of moles of oxygen
To find the number of moles (\(n\)), we use Avogadro's number (\(N_A = 6.022 \times 10^{23}\) molecules/mol):
\[ n = \frac{\text{Number of molecules}}{N_A} \]
Given the number of molecules is \(4.045 \times 10^{23}\):
\[ n = \frac{4.045 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.672 \, moles \]

### Step 4: Use the Ideal Gas Law to calculate the volume
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \(P\) = pressure in atm
- \(V\) = volume in liters
- \(n\) = number of moles
- \(R\) = ideal gas constant = \(0.0821 \, L \cdot atm/(K \cdot mol)\)
- \(T\) = temperature in Kelvin

Rearranging the equation to solve for \(V\):
\[ V = \frac{nRT}{P} \]

Substituting the values:
\[ V = \frac{(0.672 \, moles) \times (0.0821 \, L \cdot atm/(K \cdot mol)) \times (300 \, K)}{0.9211 \, atm} \]

Calculating the volume:
\[ V \approx \frac{(0.672) \times (0.0821) \times (300)}{0.9211} \approx 17.97 \, L \]

### Final Answer
The volume occupied by \(4.045 \times 10^{23}\) molecules of oxygen at \(27^\circ C\) and 700 torr is approximately \(17.97 \, L\).

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