Metallic magensium has a hexagoanl close-packed structure and a density of `1.74 g //cm^(3)`. Assume magnesium atoms to be sphere of radius r. 74.1 % of the space is occupied by atoms. Calculate the volume of each atom and the atomic radius r.(Mg = 24.31)

Answer

To solve the problem step by step, we will calculate the volume of each magnesium atom and its atomic radius based on the given data.

### Step 1: Calculate the Volume of the Unit Cell
We know that the density (D) of magnesium is given as \(1.74 \, \text{g/cm}^3\), and the molar mass (M) of magnesium is \(24.31 \, \text{g/mol}\). The number of atoms per unit cell (Z) for hexagonal close-packed (HCP) structure is 6.

Using the formula for density:
\[ D = \frac{Z \cdot M}{N_A \cdot V} \]
where:
- \(D\) = density
- \(Z\) = number of atoms in the unit cell
- \(M\) = molar mass
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(V\) = volume of the unit cell

Rearranging the formula to find the volume of the unit cell:
\[ V = \frac{Z \cdot M}{D \cdot N_A} \]

Substituting the values:
\[ V = \frac{6 \cdot 24.31 \, \text{g/mol}}{1.74 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

Calculating the volume:
\[ V = \frac{145.86}{1.74 \cdot 6.022 \times 10^{23}} = \frac{145.86}{10.48788 \times 10^{23}} \approx 1.374 \times 10^{-22} \, \text{cm}^3 \]

### Step 2: Calculate the Volume Occupied by Atoms
Given that 74.1% of the space is occupied by atoms, we calculate the volume occupied by the atoms:
\[ \text{Volume occupied by atoms} = 0.741 \cdot V \]
\[ = 0.741 \cdot 1.374 \times 10^{-22} \approx 1.018 \times 10^{-22} \, \text{cm}^3 \]

### Step 3: Calculate the Volume of One Magnesium Atom
Since the volume calculated is for 6 magnesium atoms, we find the volume of one atom:
\[ \text{Volume of one atom} = \frac{1.018 \times 10^{-22}}{6} \approx 1.697 \times 10^{-23} \, \text{cm}^3 \]

### Step 4: Calculate the Atomic Radius
The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Setting this equal to the volume of one magnesium atom:
\[ \frac{4}{3} \pi r^3 = 1.697 \times 10^{-23} \]

Solving for \(r^3\):
\[ r^3 = \frac{1.697 \times 10^{-23} \cdot 3}{4 \pi} \approx \frac{5.091 \times 10^{-23}}{12.566} \approx 4.058 \times 10^{-24} \]

Now, taking the cube root to find \(r\):
\[ r \approx (4.058 \times 10^{-24})^{1/3} \approx 1.594 \times 10^{-8} \, \text{cm} \]

### Final Results
- Volume of each magnesium atom: \(1.697 \times 10^{-23} \, \text{cm}^3\)
- Atomic radius \(r\): \(1.594 \times 10^{-8} \, \text{cm}\) or \(1.594 \, \text{Å}\)