Calculate the radius of He atoms if its van der Waal's constant 'b' is 24mL `"mol"^(-1)`. (Note: mL=cubic centimeter)

Options

A. `1.355 Å`
B. `1.314Å`
C. `1.255 Å`
D. `0.355Å`

Answer

To calculate the radius of helium (He) atoms using its van der Waals constant 'b', we can follow these steps:

### Step 1: Understand the relationship
The van der Waals constant 'b' is related to the volume occupied by one mole of atoms. The formula for 'b' is given by:

\[ b = \frac{4}{3} \pi r^3 \cdot N_A \]

where:
- \( b \) = van der Waals constant (in mL/mol)
- \( r \) = radius of the atom (in cm)
- \( N_A \) = Avogadro's number (\( 6.023 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Convert units
Since \( b \) is given in mL/mol, we can convert it to cm\(^3\)/mol (1 mL = 1 cm\(^3\)):

\[ b = 24 \, \text{cm}^3/\text{mol} \]

### Step 3: Substitute known values into the equation
Now, substitute the values into the equation:

\[ 24 = \frac{4}{3} \pi r^3 \cdot 6.023 \times 10^{23} \]

### Step 4: Solve for \( r^3 \)
Rearranging the equation to solve for \( r^3 \):

\[ r^3 = \frac{24}{\frac{4}{3} \pi \cdot 6.023 \times 10^{23}} \]

Calculating the denominator:

\[ \frac{4}{3} \pi \approx 4.18879 \]
\[ \frac{4}{3} \pi \cdot 6.023 \times 10^{23} \approx 25.189 \times 10^{23} \]

Now substituting this back:

\[ r^3 = \frac{24}{25.189 \times 10^{23}} \approx 9.52 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Calculate \( r \)
Now take the cube root to find \( r \):

\[ r \approx (9.52 \times 10^{-23})^{1/3} \approx 4.64 \times 10^{-8} \, \text{cm} \]

### Step 6: Convert to Angstroms
Since \( 1 \, \text{Å} = 10^{-8} \, \text{cm} \):

\[ r \approx 4.64 \, \text{Å} \]

### Final Answer
Thus, the radius of helium atoms is approximately:

\[ \text{Radius of He atoms} \approx 1.355 \, \text{Å} \]