Question

A metal forms fcc unit cell with unit edge length \(400\ \text{pm}\). Calculate the number of atoms present in \(0.6\ \text{g}\) of metal if the density of metal is \(12.0\ \text{g/cm}^{3}\)

• A. \(2.720\times 10^{21}\)
• B. \(3.125\times 10^{21}\)
• C. \(2.235\times 10^{21}\)
• D. \(3.650\times 10^{21}\)

Hint:

Find the mass of one unit cell from density and edge length, then count the unit cells in 0.6 g and multiply by 4 atoms.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Mass of a unit cell = density \(\times\) volume. An fcc unit cell has 4 atoms.

Step 2: Volume and mass of one cell
\(a=400\ \text{pm}=4\times 10^{-8}\ \text{cm}\)
\(V=a^{3}=6.4\times 10^{-23}\ \text{cm}^{3}\)
Mass \(=12.0\times 6.4\times 10^{-23}=7.68\times 10^{-22}\ \text{g}\)

Step 3: Unit cells in 0.6 g
\(\dfrac{0.6}{7.68\times 10^{-22}}=7.8125\times 10^{20}\)

Step 4: Number of atoms
Atoms \(=4\times 7.8125\times 10^{20}=3.125\times 10^{21}\)

Final Answer:
The number of atoms in 0.6 g is 3.125 times 10 to the power 21. \[ \boxed{\text{(B) }3.125\times 10^{21}} \]