Question

A metal crystallizes in a bcc unit cell having unit cell volume \(2.5\times 10^{-23} \text{cm}^3\). Find the number of unit cells in 18 g metal if the density of metal is \(7.2 \text{g cm}^{-3}\).

• A. \(1.0\times 10^{23}\)
• B. \(2.0\times 10^{23}\)
• C. \(0.5\times 10^{23}\)
• D. \(1.5\times 10^{23}\)

Hint:

Find the volume of the sample from the density, then divide by the volume of one unit cell.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The volume of the whole sample is mass divided by density. Dividing this by the volume of a single unit cell gives the number of unit cells. The bcc detail is not needed for this count.

Step 2: Key Formula or Approach
\[ V=\frac{m}{\rho},\qquad N_{cells}=\frac{V}{V_{cell}} \]

Step 3: Volume of the sample
\[ V=\frac{18}{7.2}=2.5\ \text{cm}^3 \]

Step 4: Number of unit cells
\[ N=\frac{2.5}{2.5\times10^{-23}}=1.0\times10^{23} \]

Step 5: Why this works
Unit cells fill space with no gaps, so the sample volume is the number of cells times the volume of one cell. The bcc fact only matters if we want atoms, which would be twice this number, \(2.0\times10^{23}\), option (B).

Final Answer:
The sample contains \(1.0\times10^{23}\) unit cells, option (A). \[ \boxed{1.0\times10^{23}} \]