van der Waals constant b of helium is 24 mL `mol^(-1)`. Find molecular diameter of helium.

Options

A. `1.335xx10^(-10)"cm" `
B. `1.335xx10^(-8)"cm" `
C. `2.67xx10^(-8)"cm" `
D. `4.34xx10^(-8)"cm" `

Explanation

To find the molecular diameter of helium using the van der Waals constant \( b \), we can follow these steps:

### Step 1: Understand the relationship of \( b \) with molecular volume
The van der Waals constant \( b \) is related to the volume occupied by one mole of gas molecules. It can be expressed as:
\[ b = 4 \times V_{\text{molecule}} \times N_A \]
where \( V_{\text{molecule}} \) is the volume occupied by one molecule and \( N_A \) is Avogadro's number.

### Step 2: Express the volume of one molecule
The volume occupied by one molecule can be modeled as the volume of a sphere:
\[ V_{\text{molecule}} = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the molecule.

### Step 3: Substitute \( V_{\text{molecule}} \) into the equation for \( b \)
Substituting the expression for \( V_{\text{molecule}} \) into the equation for \( b \):
\[ b = 4 \times \left(\frac{4}{3} \pi r^3\right) \times N_A \]
This simplifies to:
\[ b = \frac{16}{3} \pi r^3 N_A \]

### Step 4: Solve for \( r \)
Rearranging the equation to solve for \( r^3 \):
\[ r^3 = \frac{3b}{16 \pi N_A} \]
Taking the cube root gives:
\[ r = \left(\frac{3b}{16 \pi N_A}\right)^{1/3} \]

### Step 5: Substitute the known values
Given:
- \( b = 24 \, \text{mL/mol} = 24 \times 10^{-3} \, \text{L/mol} = 24 \times 10^{-6} \, \text{m}^3/\text{mol} \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( \pi \approx 3.14 \)

Substituting these values into the equation for \( r \):
\[ r = \left(\frac{3 \times (24 \times 10^{-6})}{16 \times 3.14 \times (6.022 \times 10^{23})}\right)^{1/3} \]

### Step 6: Calculate \( r \)
Calculating the value inside the parentheses:
\[ r = \left(\frac{72 \times 10^{-6}}{16 \times 3.14 \times 6.022 \times 10^{23}}\right)^{1/3} \]
Calculating the denominator:
\[ 16 \times 3.14 \times 6.022 \approx 303.84 \]
Thus:
\[ r \approx \left(\frac{72 \times 10^{-6}}{303.84 \times 10^{23}}\right)^{1/3} \]
Calculating this gives:
\[ r \approx 2.67 \times 10^{-8} \, \text{m} \]

### Step 7: Find the diameter
The diameter \( d \) is twice the radius:
\[ d = 2r \approx 2 \times 2.67 \times 10^{-8} \approx 5.34 \times 10^{-8} \, \text{m} = 5.34 \, \text{Å} \]

### Final Answer
The molecular diameter of helium is approximately \( 5.34 \, \text{Å} \).

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