Each edge of a cubic unit cell is 400pm long. If atomic mass of the elements is 120 and its desity is `6.25g//cm^(2)`, the crystal lattice is: `(use N_(A)=6 xx 10^(23))`

Options

A. primitive
B. body centred
C. Face centred
D. end centred

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the edge length of the cubic unit cell
The edge length of the cubic unit cell is given as 400 pm (picometers). We need to convert this into centimeters for consistency with the density units.

\[ \text{Edge length} = 400 \text{ pm} = 400 \times 10^{-12} \text{ m} = 4 \times 10^{-10} \text{ m} \]

Since \(1 \text{ m} = 100 \text{ cm}\),
\[ \text{Edge length} = 4 \times 10^{-10} \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 4 \times 10^{-8} \text{ cm} \]

### Step 2: Calculate the volume of the cubic unit cell
The volume \(V\) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
where \(a\) is the edge length.

\[ V = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3 \]

### Step 3: Use the density formula to find Z
The density formula relates density (\(d\)), molar mass (\(M\)), volume (\(V\)), and the number of formula units per unit cell (\(Z\)):
\[ d = \frac{Z \cdot M}{V \cdot N_A} \]
where \(N_A\) is Avogadro's number.

Rearranging the formula to solve for \(Z\):
\[ Z = \frac{d \cdot V \cdot N_A}{M} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
- Density \(d = 6.25 \text{ g/cm}^3\)
- Volume \(V = 64 \times 10^{-24} \text{ cm}^3\)
- Molar mass \(M = 120 \text{ g/mol}\)
- Avogadro's number \(N_A = 6 \times 10^{23} \text{ mol}^{-1}\)

\[ Z = \frac{6.25 \text{ g/cm}^3 \times 64 \times 10^{-24} \text{ cm}^3 \times 6 \times 10^{23} \text{ mol}^{-1}}{120 \text{ g/mol}} \]

### Step 5: Calculate \(Z\)
Calculating the numerator:
\[ 6.25 \times 64 \times 6 = 2400 \]
Now, substituting this back into the equation:
\[ Z = \frac{2400 \times 10^{-24} \text{ g/cm}^3 \cdot \text{cm}^3 \cdot \text{mol}^{-1}}{120 \text{ g/mol}} = \frac{2400}{120} = 20 \]

### Step 6: Final calculation
Now, we need to divide by \(10^{-24}\):
\[ Z = \frac{2400}{120} = 20 \Rightarrow Z = 2 \]

### Step 7: Identify the crystal lattice type
Since \(Z = 2\), this indicates that the crystal structure is a Body-Centered Cubic (BCC) lattice.

### Final Answer
The crystal lattice is **Body-Centered Cubic (BCC)**.

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