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To solve this problem and find the density of Freon-11 (CFCl₃) under the given conditions (163.0 °C and 2.74 atm), we can apply the ideal gas law, rearranged to solve for density.

### Formula:

The ideal gas law is:

\[
PV = nRT
\]

Where:
- \( P \) is pressure,
- \( V \) is volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is temperature in Kelvin.

We can express density (\( \rho \)) using:

\[
\rho = \frac{m}{V} = \frac{PM}{RT}
\]

Where:
- \( m \) is mass,
- \( M \) is the molar mass of the gas.

### Step-by-step approach:
1. **Convert the temperature to Kelvin**:
   \[
   T(K) = 163.0^\circ C + 273.15 = 436.15 \, K
   \]

2. **Use the appropriate gas constant**:
   - \( R = 0.0821 \, \text{L·atm/(mol·K)} \)

3. **Find the molar mass of Freon-11 (CFCl₃)**:
   - Molar mass \( M = 1(12.01) + 1(19.00) + 3(35.45) \, \text{g/mol} = 137.37 \, \text{g/mol} \)

4. **Substitute into the density formula**:
   \[
   \rho = \frac{PM}{RT} = \frac{(2.74 \, \text{atm}) \cdot (137.37 \, \text{g/mol})}{(0.0821 \, \text{L·atm/(mol·K}) \cdot (436.15 \, \text{K})}
   \]

Let's calculate the final result.It seems there was an issue calculating the result. I'll walk you through the remaining steps to manually calculate the density.

Using the formula:

\[
\rho = \frac{(2.74 \

Find the density of Freon-11 (CFCCl₃) at 163.0 °C and 2.74 atm. Round your answer to 3 significant figures.

Learn how to calculate the density of Freon-11 (CFCCl₃) at 163.0°C and 2.74 atm using the ideal gas law. This solution walks through the step-by-step process of using the formula ρ = PM / RT and includes the molar mass of Freon-11.

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