Question

Calculate the number of unit cells present in \(1\) g metal if the product of the density of metal and the volume of unit cell is \(2.5\times 10^{-22}\) g.

• A. \(1.0\times 10^{21}\)
• B. \(2.0\times 10^{21}\)
• C. \(3.0\times 10^{21}\)
• D. \(4.0\times 10^{21}\)

Hint:

Density times cell volume is the mass of one cell. Divide 1 g by it.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The product \(\rho \times V\) of density and unit-cell volume is the mass of one unit cell. The number of unit cells in a given mass is then total mass divided by this cell mass.

Step 2: Mass of one unit cell:
\[ m_{cell} = \rho\, a^3 = 2.5\times 10^{-22}\ \text{g} \]

Step 3: Number of cells in 1 g:
\[ N = \frac{1\ \text{g}}{2.5\times 10^{-22}\ \text{g}} = 0.4\times 10^{22} = 4.0\times 10^{21} \]

Step 4: Check the options:
\(1.0\times 10^{21}\), \(2.0\times 10^{21}\) and \(3.0\times 10^{21}\) would need cell masses of \(10^{-21}\), \(5\times 10^{-22}\) and \(3.3\times 10^{-22}\) g, none of which is given. So the answer is (D).

Final Answer:
The metal has 4.0e21 unit cells per gram, option (D). \[ \boxed{4.0\times 10^{21}} \]