`KCl` has `NaCl` type face centred cubic crystal structure and `CsF` has `CsCl` type cubic crystal structure. Calculate the ratio of densities of `CsF` and `KCl` it is given that the molar mass of `CsF` is two that of `KCl` and edge length of `KCl` unit cell of `1.5` times that for `CsF`.

Options

A. `1.68`
B. `2.72`
C. `3.12`
D. `4.62`

Explanation

To calculate the ratio of densities of `CsF` and `KCl`, we can follow these steps:

### Step 1: Understand the given data
- Molar mass of `CsF` is twice that of `KCl`.
- Edge length of `KCl` unit cell is 1.5 times that of `CsF`.
- `KCl` has a face-centered cubic (FCC) structure, which means it has 4 formula units per unit cell (Z = 4).
- `CsF` has a body-centered cubic (BCC) structure, which means it has 2 formula units per unit cell (Z = 2).

### Step 2: Write the formula for density
The density (ρ) of a substance can be calculated using the formula:
\[ \rho = \frac{Z \times M}{N_A \times a^3} \]
where:
- \(Z\) = number of formula units per unit cell,
- \(M\) = molar mass,
- \(N_A\) = Avogadro's number,
- \(a\) = edge length of the unit cell.

### Step 3: Calculate the density of `KCl`
For `KCl`:
- \(Z_{KCl} = 4\)
- Let \(M_{KCl} = M\) (molar mass of KCl)
- Edge length of `KCl` = \(a_{KCl} = 1.5 \times a_{CsF}\)

Thus, the density of `KCl` is:
\[ \rho_{KCl} = \frac{4 \times M}{N_A \times (1.5 \times a_{CsF})^3} \]
\[ = \frac{4 \times M}{N_A \times 3.375 \times a_{CsF}^3} \]
\[ = \frac{4M}{3.375 N_A a_{CsF}^3} \]

### Step 4: Calculate the density of `CsF`
For `CsF`:
- \(Z_{CsF} = 2\)
- Molar mass of `CsF` = \(2M\)

Thus, the density of `CsF` is:
\[ \rho_{CsF} = \frac{2 \times (2M)}{N_A \times a_{CsF}^3} \]
\[ = \frac{4M}{N_A a_{CsF}^3} \]

### Step 5: Calculate the ratio of densities
Now, we can find the ratio of densities:
\[ \frac{\rho_{CsF}}{\rho_{KCl}} = \frac{\frac{4M}{N_A a_{CsF}^3}}{\frac{4M}{3.375 N_A a_{CsF}^3}} \]
The \(4M\) and \(N_A a_{CsF}^3\) cancel out:
\[ = \frac{1}{\frac{1}{3.375}} = 3.375 \]

### Step 6: Final ratio
Thus, the ratio of densities of `CsF` to `KCl` is:
\[ \frac{\rho_{CsF}}{\rho_{KCl}} = 3.375 \]