Step 1: Understanding the Concept:
The compressibility factor ($Z$) is a useful correction factor that quantifies the deviation of a real gas from perfect ideal gas behavior. It is mathematically defined as the ratio of the actual molar volume of a gas to the theoretical molar volume of an ideal gas at the same temperature and pressure conditions. Step 2: Key Formula or Approach:
The fundamental formula for the compressibility factor is:
\[ Z = \frac{V_{\text{real}}}{V_{\text{ideal}}} \]
where $V_{\text{real}}$ is the actual molar volume of the real gas and $V_{\text{ideal}}$ is the theoretical molar volume of an ideal gas. Step 3: Detailed Explanation:
Given values from the problem:
Compressibility factor ($Z$) = $1.05$
The conditions specified are Standard Temperature and Pressure (STP). In traditional textbook problems, the historical standard value for the molar volume of an ideal gas at STP ($0^\circ\text{C}, 1 \text{ atm}$) is commonly used:
\[ V_{\text{ideal}} = 22.4 \text{ L/mol} = 22.4 \text{ dm}^3\text{/mol} \]
Rearranging the compressibility factor formula to solve for $V_{\text{real}}$:
\[ V_{\text{real}} = Z \times V_{\text{ideal}} \]
Substitute the given values:
\[ V_{\text{real}} = 1.05 \times 22.4 \text{ dm}^3 \]
\[ V_{\text{real}} = 23.52 \text{ dm}^3 \]
This calculated result matches option C exactly. Step 4: Final Answer:
The molar volume of the real gas is $23.52\text{dm}^3$.