A metal `M` of atomic weight 54.9 has a density of `7.42 g cm^(-3)`. Calculate the volume occupied and the radius of the atom of this metal assuming it to be sphere.

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the Mass of One Atom
Given:
- Atomic weight of metal \( M \) = 54.9 g/mol
- Avogadro's number \( N_A \) = \( 6.022 \times 10^{23} \) atoms/mol

The mass of one atom can be calculated using the formula:
\[ \text{Mass of one atom} = \frac{\text{Atomic weight}}{N_A} \]
Substituting the values:
\[ \text{Mass of one atom} = \frac{54.9 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 9.12 \times 10^{-23} \, \text{g} \]

### Step 2: Calculate the Volume Occupied by One Atom
Given:
- Density of metal \( M \) = 7.42 g/cm³

Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{9.12 \times 10^{-23} \, \text{g}}{7.42 \, \text{g/cm}^3} \approx 1.23 \times 10^{-23} \, \text{cm}^3 \]

### Step 3: Calculate the Radius of the Atom
Assuming the atom is spherical, we use the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Setting the volume equal to the calculated volume:
\[ 1.23 \times 10^{-23} \, \text{cm}^3 = \frac{4}{3} \pi r^3 \]
Rearranging to solve for \( r^3 \):
\[ r^3 = \frac{1.23 \times 10^{-23} \times 3}{4 \pi} \]
Calculating \( r^3 \):
\[ r^3 \approx \frac{3.69 \times 10^{-23}}{12.566} \approx 2.94 \times 10^{-24} \, \text{cm}^3 \]
Now, taking the cube root to find \( r \):
\[ r \approx (2.94 \times 10^{-24})^{1/3} \approx 1.42 \times 10^{-8} \, \text{cm} \]

### Final Results
- Volume occupied by one atom of metal \( M \): \( 1.23 \times 10^{-23} \, \text{cm}^3 \)
- Radius of the atom: \( 1.42 \times 10^{-8} \, \text{cm} \)