Assuming that air at STP contained 80% by volume of nitrogen, the volume of air at STP that contains `4.8xx10^(23)` molecules of notrogen is

Options

A. 18L
B. 44.8 L
C. 22.4 L
D. 11.2 L

Answer

To solve the problem, we need to find the volume of air at STP that contains \(4.8 \times 10^{23}\) molecules of nitrogen, given that air is composed of 80% nitrogen by volume.

### Step-by-Step Solution:

**Step 1: Find the number of moles of nitrogen.**
To find the number of moles of nitrogen from the number of molecules, we use Avogadro's number, which is \(6.022 \times 10^{23}\) molecules per mole.

\[ \text{Number of moles of nitrogen} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{4.8 \times 10^{23}}{6.022 \times 10^{23}} \]

Calculating this gives:

\[ \text{Number of moles of nitrogen} = 0.797 \text{ moles} \]

**Step 2: Calculate the volume of nitrogen at STP.**
At STP (Standard Temperature and Pressure), one mole of any gas occupies \(22.4\) liters. Therefore, the volume of nitrogen can be calculated as follows:

\[ \text{Volume of nitrogen} = \text{Number of moles} \times \text{Volume per mole} = 0.797 \times 22.4 \text{ L} \]

Calculating this gives:

\[ \text{Volume of nitrogen} \approx 17.87 \text{ L} \]

**Step 3: Calculate the volume of air that contains this volume of nitrogen.**
Since air is composed of 80% nitrogen by volume, we can set up the equation:

\[ \text{Volume of nitrogen} = 0.8 \times \text{Volume of air} \]

Let \(V\) be the volume of air. Then we have:

\[ 17.87 = 0.8 \times V \]

To find \(V\), we rearrange the equation:

\[ V = \frac{17.87}{0.8} \]

Calculating this gives:

\[ V \approx 22.34 \text{ L} \]

**Step 4: Round the answer.**
Since the question asks for the volume of air at STP, we can round this to an appropriate significant figure:

\[ V \approx 22.4 \text{ L} \]

### Final Answer:
The volume of air at STP that contains \(4.8 \times 10^{23}\) molecules of nitrogen is approximately **22.4 liters**.

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