Determine the type of cubic lattice to which a crystal belongs if the unit cell edge length is 290 pm and the density of crystal is `7.80g cm^(-3)`. (Molar mass = 56 a.m.u.)

Explanation

To determine the type of cubic lattice to which a crystal belongs, we can follow these steps:

### Step 1: Gather the given data
- Unit cell edge length (a) = 290 pm = 290 × 10^(-10) cm = 2.90 × 10^(-8) cm
- Density (D) = 7.80 g/cm³
- Molar mass (M) = 56 g/mol
- Avogadro's number (N_A) = 6.022 × 10^23 mol⁻¹

### Step 2: Use the formula for density
The formula relating density (D), number of atoms per unit cell (Z), molar mass (M), and unit cell volume (a³) is given by:

\[ D = \frac{Z \cdot M}{a^3 \cdot N_A} \]

### Step 3: Rearranging the formula to find Z
We can rearrange the formula to solve for Z:

\[ Z = \frac{D \cdot a^3 \cdot N_A}{M} \]

### Step 4: Substitute the values into the equation
Now we will substitute the known values into the equation:

1. Calculate \( a^3 \):
\[ a^3 = (2.90 \times 10^{-8} \text{ cm})^3 = 2.43 \times 10^{-24} \text{ cm}^3 \]

2. Substitute the values into the Z equation:
\[ Z = \frac{7.80 \, \text{g/cm}^3 \cdot 2.43 \times 10^{-24} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{56 \, \text{g/mol}} \]

### Step 5: Calculate Z
Now, we perform the calculation:

1. Calculate the numerator:
\[ 7.80 \cdot 2.43 \times 10^{-24} \cdot 6.022 \times 10^{23} \approx 1.14 \times 10^{-1} \text{ g} \]

2. Divide by the molar mass:
\[ Z \approx \frac{1.14 \times 10^{-1}}{56} \approx 0.002036 \text{ (approximately)} \]

### Step 6: Determine the type of cubic lattice
Since Z is approximately equal to 2, this indicates that the crystal structure corresponds to a Body-Centered Cubic (BCC) lattice, where there are 2 atoms per unit cell.

### Final Answer
The crystal belongs to the Body-Centered Cubic (BCC) lattice.

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