If the critical constants for a hypothetical gas are `V_(c) = 150 cm^(3) mol^(-1). P_(c) = 50 atm` and `T_(c) = 300K`. then the radius of the molecule is: [Take `R = (1)/(12)Ltr atm mol^(-1)K^(-1)]`

Options

A. `((75)/(2piN_(A)))^(1//3)`
B. `((75)/(8piN_(A)))^(1//3)`
C. `((3)/(piN_(A)))^(1//3)`
D. `((3)/(256piN_(A)))^(1//3)`

Explanation

To find the radius of the molecule given the critical constants for a hypothetical gas, we can follow these steps:

### Step 1: Identify the critical constants
The critical constants provided are:
- \( V_c = 150 \, \text{cm}^3 \, \text{mol}^{-1} \)
- \( P_c = 50 \, \text{atm} \)
- \( T_c = 300 \, \text{K} \)

### Step 2: Convert the critical volume to liters
Since \( 1 \, \text{cm}^3 = 0.001 \, \text{L} \), we convert \( V_c \) to liters:
\[ V_c = 150 \, \text{cm}^3 \, \text{mol}^{-1} = 150 \times 0.001 \, \text{L} \, \text{mol}^{-1} = 0.15 \, \text{L} \, \text{mol}^{-1} \]

### Step 3: Use the relationship between critical volume and the van der Waals constant \( B \)
The relationship is given by:
\[ V_c = 3B \]
Thus, we can find \( B \):
\[ B = \frac{V_c}{3} = \frac{0.15 \, \text{L} \, \text{mol}^{-1}}{3} = 0.05 \, \text{L} \, \text{mol}^{-1} \]

### Step 4: Use the formula for the radius of the molecule
The formula for the radius \( r \) of the molecule in terms of \( B \) is:
\[ r = \left( \frac{3B}{16\pi N_A} \right)^{1/3} \]
where \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \).

### Step 5: Substitute the value of \( B \)
Substituting \( B = 0.05 \, \text{L} \, \text{mol}^{-1} \):
\[ r = \left( \frac{3 \times 0.05}{16\pi N_A} \right)^{1/3} \]

### Step 6: Calculate the radius
Substituting \( N_A \):
\[ r = \left( \frac{0.15}{16\pi (6.022 \times 10^{23})} \right)^{1/3} \]
Calculating the denominator:
\[ 16\pi \approx 50.27 \quad \text{(using } \pi \approx 3.14\text{)} \]
Thus,
\[ r = \left( \frac{0.15}{50.27 \times 6.022 \times 10^{23}} \right)^{1/3} \]
Calculating the value:
\[ r = \left( \frac{0.15}{3.025 \times 10^{25}} \right)^{1/3} \]
\[ r = \left( 4.95 \times 10^{-27} \right)^{1/3} \approx 1.71 \times 10^{-9} \, \text{m} \approx 1.71 \, \text{nm} \]

### Final Answer
The radius of the molecule is approximately \( 1.71 \, \text{nm} \).