The unit of length convenient on the atomic scale is known a an angstrom and is denoted by `A=10^(-10)m`. The radius of the hydrogen atom is about 0.5 A. What is the total atomic valume in `m^(3)` of a mole of hydrogen atoms?

Options

A. `3.15xx10^(-7)m^(3)`
B. `3.15xx10^(-9)m^(3)`
C. `3.15xx10^(-4)m^(3)`
D. `3.15xx10^(-20)m^(3)`

Answer

To find the total atomic volume of a mole of hydrogen atoms, we can follow these steps:

### Step 1: Convert the radius of the hydrogen atom from angstroms to meters.
The radius of the hydrogen atom is given as \( r = 0.5 \, \text{A} \).
Since \( 1 \, \text{A} = 10^{-10} \, \text{m} \), we can convert the radius:
\[ r = 0.5 \, \text{A} = 0.5 \times 10^{-10} \, \text{m} = 5.0 \times 10^{-11} \, \text{m} \]

### Step 2: Calculate the volume of a single hydrogen atom using the formula for the volume of a sphere.
The formula for the volume \( V \) of a sphere is:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the value of \( r \):
\[ V = \frac{4}{3} \pi (5.0 \times 10^{-11})^3 \]

### Step 3: Calculate \( (5.0 \times 10^{-11})^3 \).
\[ (5.0 \times 10^{-11})^3 = 125 \times 10^{-33} = 1.25 \times 10^{-31} \, \text{m}^3 \]

### Step 4: Substitute this value into the volume formula.
\[ V = \frac{4}{3} \pi (1.25 \times 10^{-31}) \]
Using \( \pi \approx 3.14 \):
\[ V \approx \frac{4}{3} \times 3.14 \times 1.25 \times 10^{-31} \]
Calculating this gives:
\[ V \approx \frac{4 \times 3.14 \times 1.25}{3} \times 10^{-31} \approx 5.24 \times 10^{-31} \, \text{m}^3 \]

### Step 5: Calculate the volume of one mole of hydrogen atoms.
The number of atoms in one mole is given by Avogadro's number, \( N_A = 6.022 \times 10^{23} \).
Thus, the total volume \( V_{total} \) of one mole of hydrogen atoms is:
\[ V_{total} = N_A \times V \]
Substituting the values:
\[ V_{total} = 6.022 \times 10^{23} \times 5.24 \times 10^{-31} \]

### Step 6: Perform the multiplication.
\[ V_{total} \approx 6.022 \times 5.24 \times 10^{-7} \approx 3.15 \times 10^{-7} \, \text{m}^3 \]

### Final Answer:
The total atomic volume of a mole of hydrogen atoms is approximately:
\[ \boxed{3.15 \times 10^{-7} \, \text{m}^3} \]