CsCl has body centred cubic lattice with the length of a side of a unit cell 412.1 pm and alumininum is face centred cubic lattice with length of the side of unit cell 405 pm. Which of the two has larger density ? (Atomic mass of `Cs = 132.9, Al = 26.9 , Cl = 35.5)

Explanation

To determine which of the two compounds, CsCl (Cesium Chloride) or Al (Aluminum), has a larger density, we will use the formula for density in a crystalline solid, which is given by:

\[ \rho = \frac{Z \cdot M_A}{A^3 \cdot N_A} \]

Where:
- \(\rho\) = density
- \(Z\) = number of formula units (or atoms) per unit cell
- \(M_A\) = molar mass of the substance (in grams per mole)
- \(A\) = length of the side of the unit cell (in cm)
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))

### Step 1: Calculate the density of CsCl

1. **Identify the parameters for CsCl:**
- CsCl has a body-centered cubic (BCC) structure, so \(Z = 2\) (1 Cs atom and 1 Cl atom per unit cell).
- Molar mass of CsCl = mass of Cs + mass of Cl = \(132.9 \, \text{g/mol} + 35.5 \, \text{g/mol} = 168.4 \, \text{g/mol}\).
- Side length \(A = 412.1 \, \text{pm} = 412.1 \times 10^{-10} \, \text{cm}\).

2. **Calculate the volume of the unit cell:**
\[ A^3 = (412.1 \times 10^{-10} \, \text{cm})^3 = 7.0 \times 10^{-29} \, \text{cm}^3 \]

3. **Substitute the values into the density formula:**
\[ \rho_{CsCl} = \frac{Z \cdot M_A}{A^3 \cdot N_A} = \frac{2 \cdot 168.4}{7.0 \times 10^{-29} \cdot 6.022 \times 10^{23}} \]

4. **Calculate the density:**
\[ \rho_{CsCl} = \frac{336.8}{4.22 \times 10^{-5}} \approx 7.99 \, \text{g/cm}^3 \]

### Step 2: Calculate the density of Aluminum

1. **Identify the parameters for Aluminum:**
- Aluminum has a face-centered cubic (FCC) structure, so \(Z = 4\).
- Molar mass of Al = \(26.9 \, \text{g/mol}\).
- Side length \(A = 405 \, \text{pm} = 405 \times 10^{-10} \, \text{cm}\).

2. **Calculate the volume of the unit cell:**
\[ A^3 = (405 \times 10^{-10} \, \text{cm})^3 = 6.63 \times 10^{-29} \, \text{cm}^3 \]

3. **Substitute the values into the density formula:**
\[ \rho_{Al} = \frac{Z \cdot M_A}{A^3 \cdot N_A} = \frac{4 \cdot 26.9}{6.63 \times 10^{-29} \cdot 6.022 \times 10^{23}} \]

4. **Calculate the density:**
\[ \rho_{Al} = \frac{107.6}{3.99 \times 10^{-5}} \approx 2.69 \, \text{g/cm}^3 \]

### Conclusion

Comparing the densities:
- Density of CsCl = \(7.99 \, \text{g/cm}^3\)
- Density of Al = \(2.69 \, \text{g/cm}^3\)

Thus, **CsCl has a larger density than Aluminum**.