CsBr crystallizes in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that atomic mass of $Cs=133$ and that of $Br=80~amu$ and Avogadro number being $6.023 imes 10^{23}mol^{-1}$ the density of CsBr is:

Question

CsBr crystallizes in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that atomic mass of $Cs=133$ and that of $Br=80~amu$ and Avogadro number being $6.023 	imes 10^{23}mol^{-1}$ the density of CsBr is:

Accepted Answer

To determine the density of CsBr, which crystallizes in a body-centered cubic (BCC) lattice, we need to use the formula for density ($\rho$) of a crystalline solid:

$$
\rho = \frac{Z \cdot M}{a^3 \cdot N_A}
$$

where:
- $Z$ is the number of formula units per unit cell.
- $M$ is the molar mass of the compound.
- $a$ is the edge length of the unit cell.
- $N_A$ is Avogadro's number.

### Step-by-step Calculation:

1. **Determine $Z$:**
   - In a body-centered cubic lattice, there is 1 atom at each corner of the cube and 1 atom in the center. However, for CsBr, which is an ionic compound, the Cs and Br ions occupy these positions. In a BCC structure, there is effectively 1 formula unit per unit cell ($Z = 1$).

2. **Calculate the Molar Mass ($M$):**
   - The molar mass of CsBr is the sum of the atomic masses of Cs and Br:
     $$
     M = 133 \, \text{amu} + 80 \, \text{amu} = 213 \, \text{g/mol}
     $$

3. **Convert the Unit Cell Length ($a$) to cm:**
   - Given $a = 436.6 \, \text{pm}$, convert this to cm:
     $$
     a = 436.6 \times 10^{-12} \, \text{m} = 436.6 \times 10^{-10} \, \text{cm}
     $$

4. **Use Avogadro's Number ($N_A$):**
   - $N_A = 6.023 \times 10^{23} \, \text{mol}^{-1}$

5. **Calculate the Density ($\rho$):**
   - Substitute the values into the density formula:
     $$
     \rho = \frac{1 \cdot 213}{(436.6 \times 10^{-10})^3 \cdot 6.023 \times 10^{23}}
     $$
   - Calculate $a^3$:
     $$
     a^3 = (436.6 \times 10^{-10})^3 = 8.32 \times 10^{-23} \, \text{cm}^3
     $$
   - Calculate $\rho$:
     $$
     \rho = \frac{213}{8.32 \times 10^{-23} \cdot 6.023 \times 10^{23}} = \frac{213}{5.01} \approx 4.25 \, \text{g/cm}^3
     $$

### Explanation of Options:

- **Option 1: $42.5~gcm^{-3}$** is incorrect because it is an order of magnitude too high for typical ionic compounds.
- **Option 2: $0.425~gcm^{-3}$** is incorrect because it is too low for a solid ionic compound like CsBr.
- **Option 3: $8.25~gcm^{-3}$** is incorrect because it is higher than the calculated density.
- **Option 4: $4.25~gcm^{-3}$** is correct as it matches the calculated density using the given parameters.

Thus, the correct answer is Option 4: $4.25~gcm^{-3}$.