32 grams of `O_(2)` at STP will occupy volume equal to

Options

A. 22.4 liters
B. 28 liters
C. 22400 liters
D. 2.24 liters

Answer

To solve the problem of how much volume 32 grams of \( O_2 \) will occupy at STP (Standard Temperature and Pressure), follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Molar Volume**:
At STP, one mole of any ideal gas occupies a volume of 22.4 liters.

2. **Calculate the Molar Mass of \( O_2 \)**:
The molar mass of oxygen (\( O \)) is 16 g/mol. Since \( O_2 \) has two oxygen atoms, its molar mass is:
\[ \text{Molar mass of } O_2 = 2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol} \]

3. **Calculate the Number of Moles of \( O_2 \)**:
Use the formula for the number of moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} \]

4. **Calculate the Volume Occupied by \( O_2 \)**:
Since we have 1 mole of \( O_2 \), we can find the volume it occupies at STP:
\[ \text{Volume} = \text{Number of moles} \times \text{Molar volume} = 1 \, \text{mol} \times 22.4 \, \text{L/mol} = 22.4 \, \text{L} \]

5. **Final Answer**:
Therefore, 32 grams of \( O_2 \) at STP will occupy a volume of 22.4 liters.