Crystalline CsBr has a cubic structure. Calculate the unit cell edge length if the density of CsBr crystal is `4.24 g cm^(-3)` (atomic masses : `Cs = 133, Br = 80)`.

Explanation

To calculate the unit cell edge length of crystalline CsBr, we will use the formula relating density, molar mass, and the unit cell volume. Here are the steps to solve the problem:

### Step 1: Understand the formula
The formula for density (d) is given by:
\[ d = \frac{Z \cdot M}{A^3 \cdot N_0} \]
Where:
- \(d\) = density of the crystal
- \(Z\) = number of formula units in the unit cell
- \(M\) = molar mass of the compound
- \(A\) = edge length of the unit cell
- \(N_0\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))

### Step 2: Calculate the molar mass of CsBr
The molar mass \(M\) of CsBr can be calculated as follows:
\[ M = \text{Atomic mass of Cs} + \text{Atomic mass of Br} = 133 \, \text{g/mol} + 80 \, \text{g/mol} = 213 \, \text{g/mol} \]

### Step 3: Determine the number of formula units in the unit cell (Z)
For a simple cubic structure, there is 1 formula unit per unit cell:
\[ Z = 1 \]

### Step 4: Rearrange the density formula to solve for \(A^3\)
Rearranging the density formula gives:
\[ A^3 = \frac{Z \cdot M}{d \cdot N_0} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
- \(Z = 1\)
- \(M = 213 \, \text{g/mol}\)
- \(d = 4.24 \, \text{g/cm}^3\)
- \(N_0 = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

Now, substituting these values into the equation:
\[ A^3 = \frac{1 \cdot 213 \, \text{g/mol}}{4.24 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 6: Calculate \(A^3\)
Calculating the denominator:
\[ 4.24 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} = 2.55 \times 10^{24} \, \text{g/cm}^3 \cdot \text{mol}^{-1} \]
Now, substituting back:
\[ A^3 = \frac{213}{2.55 \times 10^{24}} \approx 8.35 \times 10^{-23} \, \text{cm}^3 \]

### Step 7: Calculate the edge length \(A\)
To find \(A\), take the cube root of \(A^3\):
\[ A = (8.35 \times 10^{-23})^{1/3} \approx 4.36 \times 10^{-8} \, \text{cm} \]

### Step 8: Convert to picometers
To convert centimeters to picometers:
\[ A = 4.36 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 436.0 \, \text{pm} \]

### Final Answer
Thus, the edge length of the unit cell \(A\) is approximately:
\[ \boxed{436.0 \, \text{pm}} \]