A compound having bcc geometry has atomic mass 70. Calculate the density of the unit cell if its edge length is 290 pm.

Explanation

To calculate the density of a compound with a body-centered cubic (BCC) geometry, we can follow these steps:

### Step 1: Identify the parameters
- Atomic mass (m) = 70 g/mol
- Edge length (a) = 290 pm = 290 x 10^-10 cm
- Number of atoms per unit cell (Z) for BCC = 2
- Avogadro's number (N_A) = 6.022 x 10^23 mol^-1

### Step 2: Convert edge length to centimeters
The edge length given in picometers needs to be converted to centimeters for our calculations:
\[ a = 290 \, \text{pm} = 290 \times 10^{-10} \, \text{cm} = 2.90 \times 10^{-8} \, \text{cm} \]

### Step 3: Use the formula for density
The density (ρ) of a unit cell can be calculated using the formula:
\[ \rho = \frac{Z \cdot m}{a^3 \cdot N_A} \]
Where:
- ρ = density
- Z = number of atoms per unit cell
- m = molar mass in grams
- a = edge length in cm
- N_A = Avogadro's number

### Step 4: Substitute the values into the formula
Substituting the known values into the density formula:
\[ \rho = \frac{2 \cdot 70}{(2.90 \times 10^{-8})^3 \cdot 6.022 \times 10^{23}} \]

### Step 5: Calculate the denominator
First, calculate \( (2.90 \times 10^{-8})^3 \):
\[ (2.90 \times 10^{-8})^3 = 2.90^3 \times (10^{-8})^3 = 24.389 \times 10^{-24} \, \text{cm}^3 \]

### Step 6: Complete the density calculation
Now, substituting back into the density formula:
\[ \rho = \frac{140}{24.389 \times 10^{-24} \cdot 6.022 \times 10^{23}} \]

Calculating the denominator:
\[ 24.389 \times 10^{-24} \cdot 6.022 \times 10^{23} = 1.469 \times 10^{-1} \, \text{g} \]

Now, substituting this into the density formula:
\[ \rho = \frac{140}{1.469 \times 10^{-1}} \approx 9.530 \, \text{g/cm}^3 \]

### Final Answer
The density of the unit cell is approximately:
\[ \rho \approx 9.530 \, \text{g/cm}^3 \]