KF has NaCl structure. What is the distance between `K^(+)` and `F^(-)` in KF, if its density is 2.48 g `cm^(-3)` ?

Explanation

To find the distance between \( K^+ \) and \( F^- \) in KF, we can follow these steps:

### Step 1: Identify the Structure
KF has the same structure as NaCl, which is face-centered cubic (FCC). In an FCC lattice, the distance between the cation and anion (in this case, \( K^+ \) and \( F^- \)) is given by \( \frac{A}{2} \), where \( A \) is the edge length of the unit cell.

### Step 2: Calculate the Molecular Mass of KF
The molecular mass of KF can be calculated as follows:
- Atomic mass of \( K \) (Potassium) = 39 g/mol
- Atomic mass of \( F \) (Fluorine) = 19 g/mol

Thus, the molecular mass of KF:
\[ M = 39 + 19 = 58 \text{ g/mol} \]

### Step 3: Use the Density Formula
The density (\( d \)) of a substance is given by the formula:
\[ d = \frac{z \cdot M}{A^3 \cdot N_A} \]
where:
- \( z \) = number of formula units per unit cell (for FCC, \( z = 4 \))
- \( M \) = molar mass
- \( A \) = edge length of the unit cell
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\))

### Step 4: Rearranging the Density Formula
We can rearrange the formula to solve for \( A^3 \):
\[ A^3 = \frac{z \cdot M}{d \cdot N_A} \]

### Step 5: Substitute the Values
Now, substituting the known values into the equation:
- \( z = 4 \)
- \( M = 58 \text{ g/mol} \)
- \( d = 2.48 \text{ g/cm}^3 \)
- \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)

\[ A^3 = \frac{4 \cdot 58}{2.48 \cdot 6.022 \times 10^{23}} \]

### Step 6: Calculate \( A^3 \)
Calculating \( A^3 \):
\[ A^3 = \frac{232}{14.92576 \times 10^{23}} \approx 1.554 \times 10^{-22} \text{ cm}^3 \]

### Step 7: Find \( A \)
Now, take the cube root to find \( A \):
\[ A \approx (1.554 \times 10^{-22})^{1/3} \approx 5.37 \times 10^{-8} \text{ cm} \approx 5.37 \text{ Å} \]

### Step 8: Calculate the Distance Between \( K^+ \) and \( F^- \)
The distance between \( K^+ \) and \( F^- \) is:
\[ \text{Distance} = \frac{A}{2} = \frac{5.37 \text{ Å}}{2} \approx 2.685 \text{ Å} \]

### Final Answer
The distance between \( K^+ \) and \( F^- \) in KF is approximately \( 2.685 \text{ Å} \).

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