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)/(piN)`
B. `(100)/(piN)`
C. `(6)/(piN)`
D. `(25)/(4Npi)`

Answer

To solve the problem, we need to find \( r^3 \) (the volume occupied by a single molecule of CO\(_2\)) given the critical density of CO\(_2\) and its molar mass.

### Step-by-step Solution:

1. **Understand the relationship between density, mass, and volume**:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Here, the critical density of CO\(_2\) is given as \( 0.44 \, \text{g/cm}^3 \).

2. **Calculate the molar mass of CO\(_2\)**:
The molar mass of CO\(_2\) can be calculated as follows:
\[ \text{Molar mass of CO}_2 = 12 \, \text{g/mol (C)} + 16 \, \text{g/mol (O)} \times 2 = 44 \, \text{g/mol} \]

3. **Determine the mass of one molecule of CO\(_2\)**:
Using Avogadro's number (\( N_A \)), which is approximately \( 6.022 \times 10^{23} \, \text{molecules/mol} \):
\[ \text{Mass of one molecule} = \frac{44 \, \text{g}}{N_A} \]

4. **Express the volume occupied by one mole of CO\(_2\)**:
The volume occupied by one mole of CO\(_2\) can be expressed in terms of the critical volume:
\[ V = \frac{\text{Mass}}{\text{Density}} = \frac{44 \, \text{g}}{0.44 \, \text{g/cm}^3} = 100 \, \text{cm}^3 \]

5. **Relate the volume to the volume occupied by one molecule**:
The total volume occupied by \( N_A \) molecules is equal to the volume of one mole:
\[ V = N_A \times \text{Volume of one molecule} \]
Therefore, we can express the volume of one molecule as:
\[ \text{Volume of one molecule} = \frac{100 \, \text{cm}^3}{N_A} \]

6. **Relate the volume of one molecule to the radius**:
The volume of one molecule can also be expressed as:
\[ \text{Volume of one molecule} = \frac{4}{3} \pi r^3 \]
Setting the two expressions for the volume equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{100 \, \text{cm}^3}{N_A} \]

7. **Solve for \( r^3 \)**:
Rearranging the equation to find \( r^3 \):
\[ r^3 = \frac{100 \, \text{cm}^3 \cdot 3}{4 \pi N_A} \]
Simplifying gives:
\[ r^3 = \frac{300}{4 \pi N_A} = \frac{75}{\pi N_A} \]

### Final Result:
Thus, the approximate value of \( r^3 \) in cm\(^3\) is:
\[ r^3 \approx \frac{75}{\pi N_A} \, \text{cm}^3 \]