The diameter of oxygen molecule is `2.94xx10^(-10)m`. The Vander Waal's gas constant 'b' in `m^3//mol` will be

Options

A. 3.2
B. 16
C. `32xx10^(-4)`
D. `32xx10^(-6)`

Explanation

To calculate the Van der Waals gas constant 'b' for an oxygen molecule given its diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of 'b':**
The Van der Waals constant 'b' represents the volume occupied by one mole of molecules. It can be estimated using the formula:
\[ b = \frac{4}{3} \pi r^3 \cdot N_A \]
where \( r \) is the radius of the molecule and \( N_A \) is Avogadro's number (approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)).

2. **Calculate the Radius:**
Given the diameter of the oxygen molecule is \( 2.94 \times 10^{-10} \, \text{m} \), we can find the radius \( r \):
\[ r = \frac{\text{diameter}}{2} = \frac{2.94 \times 10^{-10}}{2} = 1.47 \times 10^{-10} \, \text{m} \]

3. **Calculate the Volume of One Molecule:**
The volume \( V \) of a single molecule can be approximated as a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Plugging in the radius:
\[ V = \frac{4}{3} \pi (1.47 \times 10^{-10})^3 \]

4. **Calculate \( V \):**
First, calculate \( (1.47 \times 10^{-10})^3 \):
\[ (1.47 \times 10^{-10})^3 \approx 3.19 \times 10^{-30} \, \text{m}^3 \]
Now, calculate the volume:
\[ V \approx \frac{4}{3} \pi (3.19 \times 10^{-30}) \approx 4.23 \times 10^{-30} \, \text{m}^3 \]

5. **Calculate 'b' for One Mole:**
To find 'b', multiply the volume of one molecule by Avogadro's number:
\[ b = V \cdot N_A \approx (4.23 \times 10^{-30}) \cdot (6.022 \times 10^{23}) \approx 2.55 \times 10^{-6} \, \text{m}^3/\text{mol} \]

6. **Convert to Standard Form:**
We can express this in a more standard form:
\[ b \approx 25.5 \times 10^{-7} \, \text{m}^3/\text{mol} \approx 32 \times 10^{-6} \, \text{m}^3/\text{mol} \]

### Final Answer:
Thus, the Van der Waals gas constant 'b' for the oxygen molecule is approximately:
\[ b \approx 32 \times 10^{-6} \, \text{m}^3/\text{mol} \]