Question

Mass of unit cell of an element is \(4.15 \times 10^{-24}\) g. If edge length of unit cell is \(3.50 \times 10^{-8}\) cm, what is the density of element?

• A. 4.67 g/cm\(^3\)
• B. 1.18 g/cm\(^3\)
• C. 7.32 g/cm\(^3\)
• D. 9.67 g/cm\(^3\)

Hint:

Density of a crystal depends directly on mass of unit cell and inversely on cube of edge length.

Answer 1

The Correct Option is D

Step 1: Write the formula for density.
\[ \text{Density} = \frac{\text{mass of unit cell}}{\text{volume of unit cell}} \]
Step 2: Calculate volume of unit cell.
\[ V = a^3 = (3.50 \times 10^{-8})^3 \]
\[ V = 4.29 \times 10^{-23}\ \text{cm}^3 \]
Step 3: Substitute values.
\[ \text{Density} = \frac{4.15 \times 10^{-24}}{4.29 \times 10^{-23}} \]
\[ \text{Density} = 9.67\ \text{g/cm}^3 \]
Step 4: Conclusion.
Density of the element is 9.67 g/cm\(^3\).