Sodium crystallises in bcc arrangement with the interfacial separation between the atoms at the edge length of 53 pm. The density of the solid is

Options

A. 1.23 g/cc
B. 485 g/cc
C. 4.85 g/cc
D. 123 g/cc

Explanation

To find the density of sodium in a body-centered cubic (BCC) arrangement, we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC unit cell, there are 2 atoms per unit cell. One atom is located at each corner of the cube (8 corners, contributing 1/8 of each atom to the unit cell) and one atom is at the center of the cube.

### Step 2: Determine the Edge Length
The interatomic distance (the distance between the centers of two atoms) in a BCC structure is equal to the edge length (a) of the cube. Given that the interfacial separation (edge length) is 53 pm (picometers), we can convert this to meters for calculation purposes:
\[ a = 53 \text{ pm} = 53 \times 10^{-12} \text{ m} \]

### Step 3: Calculate the Volume of the Unit Cell
The volume (V) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (53 \times 10^{-12})^3 = 1.48 \times 10^{-21} \text{ m}^3 \]

### Step 4: Calculate the Mass of Sodium in the Unit Cell
To find the mass of sodium in the unit cell, we need to know the molar mass of sodium (Na), which is approximately 23 g/mol. We also need to convert this to kilograms:
\[ \text{Molar mass of Na} = 23 \text{ g/mol} = 23 \times 10^{-3} \text{ kg/mol} \]
Using Avogadro's number (approximately \(6.022 \times 10^{23} \text{ atoms/mol}\)), we can find the mass of the sodium in one unit cell:
\[ \text{Mass of 2 atoms of Na} = \frac{2 \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} \times 23 \times 10^{-3} \text{ kg/mol} \]
Calculating this gives:
\[ \text{Mass} = \frac{2 \times 23 \times 10^{-3}}{6.022 \times 10^{23}} \approx 7.63 \times 10^{-25} \text{ kg} \]

### Step 5: Calculate the Density
Density (\( \rho \)) is defined as mass per unit volume:
\[ \rho = \frac{\text{mass}}{\text{volume}} \]
Substituting the values:
\[ \rho = \frac{7.63 \times 10^{-25} \text{ kg}}{1.48 \times 10^{-21} \text{ m}^3} \approx 0.0515 \text{ kg/m}^3 \]
To convert this to g/cm³ (since 1 kg/m³ = 0.001 g/cm³):
\[ \rho \approx 0.0515 \times 1000 \approx 51.5 \text{ g/cm}^3 \]

### Final Answer
The density of solid sodium in a BCC arrangement is approximately **51.5 g/cm³**.