The critical density of the gas `CO_(2)` is `0.44 g cm^(-3)` at a certain temperature. If r is the radius of the molecules, `r^(3)` in `cm^(3)` is approximately. `(N` is Avogadro number)

Options

A. `(25)/(pi N)`
B. `(100)/(pi N)`
C. `(6)/(pi N)`
D. `(25)/(4N pi)`

Explanation

To solve the problem, we need to find \( r^3 \) in cm³, given the critical density of CO₂ is \( 0.44 \, \text{g/cm}^3 \) and the molecular mass of CO₂ is \( 44 \, \text{g/mol} \).

### Step-by-Step Solution:

1. **Understand the Relationship Between Density, Mass, and Volume**:
The critical density (\( \rho_c \)) is defined as:
\[ \rho_c = \frac{\text{Critical Mass}}{\text{Critical Volume}} \]

2. **Identify the Mass of One Molecule**:
The molecular mass of CO₂ is \( 44 \, \text{g/mol} \). To find the mass of one molecule, we use Avogadro's number (\( N \)):
\[ \text{Mass of one molecule} = \frac{44 \, \text{g}}{N} \]

3. **Express Critical Volume in Terms of Molecular Volume**:
The critical volume can be expressed as:
\[ V_c = 3 \times \text{Volume of one molecule} = 3 \times \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^3 \]

4. **Set Up the Equation Using Critical Density**:
Substitute the expressions for mass and volume into the density equation:
\[ 0.44 = \frac{\frac{44}{N}}{4 \pi r^3} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 0.44 \times 4 \pi r^3 = \frac{44}{N} \]
\[ r^3 = \frac{44}{0.44 \times 4 \pi N} \]

6. **Simplifying the Expression**:
Simplifying further:
\[ r^3 = \frac{44}{1.76 \pi N} \]
\[ r^3 = \frac{100}{4 \pi N} \]
\[ r^3 = \frac{25}{\pi N} \]

### Final Result:
Thus, the value of \( r^3 \) in cm³ is:
\[ r^3 \approx \frac{25}{\pi N} \]