CsBr crystallises in a body centred cubic lattice. The unit cell length is 436 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avagadro number being `6.02xx10^(23)mol^(-1)` the density of CsBr is

Options

A. `42.5g//cm^(3)`
B. `0.425 g//cm^(3)`
C. `8.53g//cm^(3)`
D. `4.25g//cm^(3)`

Answer

To find the density of CsBr, which crystallizes in a body-centered cubic (BCC) lattice, we can use the formula for density:

\[ \text{Density} = \frac{Z \times M}{N_A \times a^3} \]

Where:
- \( Z \) = Number of atoms per unit cell
- \( M \) = Molar mass of the compound (in grams per mole)
- \( N_A \) = Avogadro's number (approximately \( 6.02 \times 10^{23} \) mol\(^{-1}\))
- \( a \) = Edge length of the unit cell (in cm)

### Step 1: Determine the Molar Mass of CsBr
The molar mass of CsBr can be calculated by adding the atomic masses of cesium (Cs) and bromine (Br):

\[ M = \text{Atomic mass of Cs} + \text{Atomic mass of Br} = 133 \, \text{g/mol} + 80 \, \text{g/mol} = 213 \, \text{g/mol} \]

### Step 2: Convert the Edge Length to Centimeters
The given edge length is in picometers (pm). We need to convert it to centimeters (cm):

\[ a = 436 \, \text{pm} = 436 \times 10^{-12} \, \text{m} = 436 \times 10^{-10} \, \text{cm} \]

### Step 3: Determine the Number of Atoms per Unit Cell (Z)
In a body-centered cubic lattice, there are:
- 1 atom at the center
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom

Thus, the total number of atoms per unit cell is:

\[ Z = 1 + 8 \times \frac{1}{8} = 1 + 1 = 2 \]

### Step 4: Calculate the Volume of the Unit Cell
The volume of the unit cell can be calculated using the edge length:

\[ \text{Volume} = a^3 = (436 \times 10^{-10} \, \text{cm})^3 \]

Calculating this gives:

\[ \text{Volume} = 436^3 \times 10^{-30} \, \text{cm}^3 = 8.36 \times 10^{-24} \, \text{cm}^3 \]

### Step 5: Substitute Values into the Density Formula
Now we can substitute the values into the density formula:

\[ \text{Density} = \frac{Z \times M}{N_A \times a^3} = \frac{2 \times 213}{6.02 \times 10^{23} \times 8.36 \times 10^{-24}} \]

Calculating this gives:

\[ \text{Density} = \frac{426}{5.04} \approx 84.5 \, \text{g/cm}^3 \]

### Step 6: Final Calculation
Now, we can compute the final density value:

\[ \text{Density} \approx 8.53 \, \text{g/cm}^3 \]

Thus, the density of CsBr is approximately **8.53 g/cm³**.

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