If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.

Options

A. 2923` kgm^-3`
B. 5846` kgm^-3`
C. 8768` kgm^-3`
D. 1750` kgm^-3`

Answer

To calculate the density of a metal that forms a Face-Centered Cubic (FCC) lattice with a unit edge length of 500 pm and an atomic mass of 110 g/mol, we can follow these steps:

### Step 1: Understand the formula for density
The density (D) of a substance can be calculated using the formula:
\[ D = \frac{Z \times M}{a^3 \times N_A} \]
where:
- \( Z \) = number of atoms per unit cell (for FCC, \( Z = 4 \))
- \( M \) = molar mass in kg/mol
- \( a \) = edge length of the unit cell in meters
- \( N_A \) = Avogadro's number (\( 6.02 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Convert the edge length from picometers to meters
Given:
\[ a = 500 \text{ pm} = 500 \times 10^{-12} \text{ m} \]

### Step 3: Convert the atomic mass from grams per mole to kilograms per mole
Given:
\[ M = 110 \text{ g/mol} = 110 \times 10^{-3} \text{ kg/mol} \]

### Step 4: Substitute the values into the density formula
Now we can substitute the values into the density formula:
- \( Z = 4 \)
- \( M = 110 \times 10^{-3} \text{ kg/mol} \)
- \( a = 500 \times 10^{-12} \text{ m} \)
- \( N_A = 6.02 \times 10^{23} \text{ mol}^{-1} \)

The formula becomes:
\[ D = \frac{4 \times (110 \times 10^{-3})}{(500 \times 10^{-12})^3 \times (6.02 \times 10^{23})} \]

### Step 5: Calculate \( a^3 \)
Calculate \( a^3 \):
\[ a^3 = (500 \times 10^{-12})^3 = 125 \times 10^{-36} \text{ m}^3 \]

### Step 6: Substitute \( a^3 \) into the density formula
Now substitute \( a^3 \) back into the density formula:
\[ D = \frac{4 \times (110 \times 10^{-3})}{(125 \times 10^{-36}) \times (6.02 \times 10^{23})} \]

### Step 7: Calculate the denominator
Calculate the denominator:
\[ (125 \times 10^{-36}) \times (6.02 \times 10^{23}) = 7.525 \times 10^{-13} \]

### Step 8: Calculate the density
Now calculate the density:
\[ D = \frac{4 \times (110 \times 10^{-3})}{7.525 \times 10^{-13}} \]
\[ D = \frac{440 \times 10^{-3}}{7.525 \times 10^{-13}} \]
\[ D \approx 5846 \text{ kg/m}^3 \]

### Final Answer
The density of the metal is approximately \( 5846 \text{ kg/m}^3 \).

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