KBr has NaCl type sturcture . Its density is 2.75 g `cm^(-3)` . Edge length of unit cell will be:
(Molar mass of KBr =119 g `"mol"^(-1)`)

Options

A. `3.3xx10^(-8)` cm
B. `6.6 xx 10^(-8)` cm
C. `9.9xx10^(-8)` cm
D. `1.6xx10^(-8)` cm

Explanation

To find the edge length of the unit cell for KBr, we can use the formula for density in relation to the unit cell parameters. Here’s a step-by-step solution:

### Step 1: Understand the Structure
KBr has an NaCl type structure, which means it has a face-centered cubic (FCC) lattice. In this type of structure, the number of formula units (Z) per unit cell is 4.

### Step 2: Write the Density Formula
The formula for density (ρ) in terms of the unit cell parameters is given by:
\[ \rho = \frac{Z \cdot M}{V \cdot N_A} \]
where:
- \( Z \) = number of formula units per unit cell (for NaCl type structure, Z = 4)
- \( M \) = molar mass of KBr (given as 119 g/mol)
- \( V \) = volume of the unit cell (which is \( A^3 \), where \( A \) is the edge length)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \))

### Step 3: Substitute the Known Values
We can rearrange the density formula to solve for the edge length \( A \):
\[ \rho = \frac{Z \cdot M}{A^3 \cdot N_A} \]
Rearranging gives:
\[ A^3 = \frac{Z \cdot M}{\rho \cdot N_A} \]
Thus,
\[ A = \left( \frac{Z \cdot M}{\rho \cdot N_A} \right)^{1/3} \]

### Step 4: Plug in the Values
Now, substitute the known values into the equation:
- \( Z = 4 \)
- \( M = 119 \, \text{g/mol} \)
- \( \rho = 2.75 \, \text{g/cm}^3 \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Calculating \( A \):
\[ A = \left( \frac{4 \cdot 119}{2.75 \cdot 6.022 \times 10^{23}} \right)^{1/3} \]

### Step 5: Calculate the Volume
First, calculate the numerator:
\[ 4 \cdot 119 = 476 \]

Now calculate the denominator:
\[ 2.75 \cdot 6.022 \times 10^{23} \approx 1.655 \times 10^{24} \]

Now, substituting these values:
\[ A = \left( \frac{476}{1.655 \times 10^{24}} \right)^{1/3} \]

### Step 6: Final Calculation
Calculating the fraction:
\[ \frac{476}{1.655 \times 10^{24}} \approx 2.88 \times 10^{-22} \]

Now take the cube root:
\[ A \approx (2.88 \times 10^{-22})^{1/3} \approx 6.6 \times 10^{-8} \, \text{cm} \]

### Conclusion
The edge length of the unit cell for KBr is approximately \( 6.6 \times 10^{-8} \, \text{cm} \).

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