The van der Waals' constant 'b' of a gas is `4pixx 10^(-4)L//mol.` How near can the centeres of the two molecules approach each other? [Use :`N_(A)=6xx10^(23)`]

Options

A. `10^(-7)` m
B. `10^(-10)` m
C. `5xx10^(-11)` m
D. `5xx10^(-9)` m

Explanation

To solve the problem, we need to find how near the centers of two gas molecules can approach each other based on the van der Waals constant 'b'. The van der Waals constant 'b' is related to the volume occupied by one mole of the gas and can be expressed in terms of the radius of the molecules.

### Step-by-Step Solution:

1. **Understand the Relationship**: The van der Waals constant 'b' is given by the formula:
\[ b = \frac{4}{3} \pi r^3 N_A \]
where \( r \) is the radius of one molecule and \( N_A \) is Avogadro's number.

2. **Given Values**:
- The value of \( b \) is given as \( 4 \pi \times 10^{-4} \, \text{L/mol} \).
- Avogadro's number \( N_A = 6 \times 10^{23} \, \text{molecules/mol} \).

3. **Rearranging the Formula**: We need to find \( r \), so we rearrange the formula:
\[ r^3 = \frac{3b}{4\pi N_A} \]

4. **Substituting the Values**: Substitute the known values into the rearranged formula:
\[ r^3 = \frac{3 \times (4 \pi \times 10^{-4})}{4\pi \times (6 \times 10^{23})} \]
The \( 4\pi \) cancels out:
\[ r^3 = \frac{3 \times 10^{-4}}{6 \times 10^{23}} = \frac{0.5 \times 10^{-4}}{10^{23}} = 0.5 \times 10^{-27} \, \text{cm}^3 \]

5. **Calculating \( r \)**: Now we take the cube root to find \( r \):
\[ r = \sqrt[3]{0.5 \times 10^{-27}} \approx 5 \times 10^{-9} \, \text{cm} \]

6. **Finding the Closest Distance Between Molecules**: The closest distance between the centers of two molecules is \( 2r \):
\[ 2r = 2 \times (5 \times 10^{-9}) = 10 \times 10^{-9} \, \text{cm} = 10^{-8} \, \text{cm} \]

7. **Convert to Meters**: To convert centimeters to meters:
\[ 10^{-8} \, \text{cm} = 10^{-10} \, \text{m} \]

### Final Answer:
The centers of the two molecules can approach each other to a distance of \( 10^{-10} \, \text{m} \).