Calculate molecular diameter for a gas if its molar exclued volume is `3.2 pi ml`. (in nenometer) (Take `N_(A) = 6.0 xx 10^(23))`

Explanation

To calculate the molecular diameter for a gas given its molar excluded volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Molar Excluded Volume**:
The molar excluded volume (V_ex) is the volume that is not available for the movement of gas molecules due to their finite size. It is given as \( V_{ex} = 3.2 \pi \, \text{ml} \).

2. **Convert Molar Excluded Volume to Liters**:
Since \( 1 \, \text{ml} = 10^{-3} \, \text{L} \), we convert the molar excluded volume:
\[ V_{ex} = 3.2 \pi \, \text{ml} = 3.2 \pi \times 10^{-3} \, \text{L} \]

3. **Use the Formula for Molar Excluded Volume**:
The molar excluded volume can be expressed in terms of the radius (r) of the gas molecules:
\[ V_{ex} = N_A \times 4 \times \frac{4}{3} \pi r^3 \]
where \( N_A \) is Avogadro's number \( (6.0 \times 10^{23}) \).

4. **Set Up the Equation**:
Plugging in the values, we have:
\[ 3.2 \pi \times 10^{-3} = 6.0 \times 10^{23} \times 4 \times \frac{4}{3} \pi r^3 \]

5. **Cancel Out π**:
Dividing both sides by \( \pi \):
\[ 3.2 \times 10^{-3} = 6.0 \times 10^{23} \times \frac{16}{3} r^3 \]

6. **Rearranging the Equation**:
Rearranging to solve for \( r^3 \):
\[ r^3 = \frac{3.2 \times 10^{-3} \times 3}{6.0 \times 10^{23} \times 16} \]

7. **Calculate the Value**:
Calculate the right side:
\[ r^3 = \frac{9.6 \times 10^{-3}}{96.0 \times 10^{23}} = \frac{9.6}{96.0} \times 10^{-3 - 23} = 0.1 \times 10^{-26} = 10^{-27} \, \text{m}^3 \]

8. **Find the Radius**:
Taking the cube root:
\[ r = (10^{-27})^{1/3} = 10^{-9} \, \text{m} \]

9. **Calculate the Diameter**:
The diameter \( d \) is twice the radius:
\[ d = 2r = 2 \times 10^{-9} \, \text{m} = 2 \, \text{nm} \]

### Final Answer:
The molecular diameter for the gas is **2 nanometers**.