CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being `6.02 xx 10^23 "mol"^(-1)` the density of CsBr is

Options

A. `0.425 g//cm^3`
B. `8.5 g//cm^3`
C. `4.25 g//cm^3`
D. `82.5 g//cm^3`

Explanation

To find the density of CsBr (cesium bromide) which crystallizes in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Identify the parameters
- **Unit cell type**: Body-centered cubic (BCC)
- **Unit cell length (A)**: 436.6 pm = 436.6 x 10^-10 m
- **Atomic mass of Cs**: 133 amu
- **Atomic mass of Br**: 80 amu
- **Avogadro's number (N₀)**: 6.02 x 10^23 mol⁻¹

### Step 2: Calculate the total number of atoms in the unit cell (z)
In a body-centered cubic lattice, there are:
- 8 corner atoms (each contributes 1/8 to the unit cell)
- 1 atom at the body center

Thus, the total number of atoms (z) in the unit cell is:
\[ z = 8 \times \frac{1}{8} + 1 = 2 \]

### Step 3: Calculate the molar mass (M) of CsBr
The molar mass of CsBr is the sum of the atomic masses of Cs and Br:
\[ M = 133 \, \text{g/mol} + 80 \, \text{g/mol} = 213 \, \text{g/mol} \]

### Step 4: Calculate the volume of the unit cell (V)
The volume of the cubic unit cell is given by:
\[ V = A^3 \]
Where A is the edge length in cm. First, convert A from pm to cm:
\[ A = 436.6 \, \text{pm} = 436.6 \times 10^{-10} \, \text{m} = 4.366 \times 10^{-8} \, \text{cm} \]
Now calculate the volume:
\[ V = (4.366 \times 10^{-8} \, \text{cm})^3 \]
\[ V = 8.34 \times 10^{-24} \, \text{cm}^3 \]

### Step 5: Calculate the density (ρ)
The density can be calculated using the formula:
\[ \rho = \frac{z \times M}{V \times N₀} \]
Substituting the values:
\[ \rho = \frac{2 \times 213 \, \text{g/mol}}{8.34 \times 10^{-24} \, \text{cm}^3 \times 6.02 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 6: Solve for density
Calculating the denominator:
\[ V \times N₀ = 8.34 \times 10^{-24} \, \text{cm}^3 \times 6.02 \times 10^{23} \, \text{mol}^{-1} \approx 5.02 \times 10^{-1} \, \text{g} \]
Now substituting back into the density formula:
\[ \rho = \frac{2 \times 213}{5.02 \times 10^{-1}} \]
\[ \rho \approx \frac{426}{0.502} \approx 848.41 \, \text{g/cm}^3 \]

### Final Answer:
The density of CsBr is approximately **8.5 g/cm³**.

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