Question

How many atoms of niobium are present in $2.43\ \text{g}$ if it forms bcc structure with density $9\ \text{g}\ \text{cm}^{-3}$ and volume of unit cell $2.7 \times 10^{-23}\ \text{cm}^3$?

• A. $3.01 \times 10^{23}$
• B. $4.1 \times 10^{22}$
• C. $5.0 \times 10^{22}$
• D. $2.0 \times 10^{22}$

Hint:

To handle scientific notation accurately, convert all decimals into standard powers of 10 right away. Turning $0.27 / (2.7 \times 10^{-23})$ into $(2.7 \times 10^{-1}) / (2.7 \times 10^{-23})$ makes the division clear and easy to solve.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the total mass of a sample of niobium, its density, and the volume of a single unit cell. We need to determine the total number of niobium atoms present in this sample given that it crystallizes in a body-centered cubic (bcc) lattice.

Step 2: Key Formula or Approach:
1. First, calculate the total volume occupied by the sample of niobium using the density formula: $$\text{Total Volume} = \frac{\text{Mass}}{\text{Density}}$$ 2. Next, find the total number of unit cells contained in this volume: $$\text{Number of Unit Cells} = \frac{\text{Total Volume}}{\text{Volume of a Single Unit Cell}}$$ 3. Finally, since a bcc unit cell contains exactly 2 constituent atoms per unit cell, multiply the number of unit cells by 2 to get the total number of atoms.

Step 3: Detailed Explanation:
Given values: Mass of niobium $= 2.43\ \text{g}$ Density $(\rho) = 9\ \text{g}\ \text{cm}^{-3}$ Volume of a single unit cell $(V_{cell}) = 2.7 \times 10^{-23}\ \text{cm}^3$ Let's compute the total volume: $$\text{Total Volume} = \frac{2.43\ \text{g}}{9\ \text{g}\ \text{cm}^{-3}} = 0.27\ \text{cm}^3$$ Now, find the number of unit cells: $$\text{Number of Unit Cells} = \frac{0.27}{2.7 \times 10^{-23}} = \frac{2.7 \times 10^{-1}}{2.7 \times 10^{-23}} = 10^{22}\ \text{unit cells}$$ Since it is a body-centered cubic (bcc) crystal structure, the number of atoms per unit cell $(Z)$ is equal to 2. $$\text{Total Atoms} = Z \times \text{Number of Unit Cells}$$ $$\text{Total Atoms} = 2 \times 10^{22} = 2.0 \times 10^{22}\ \text{atoms}$$

Step 4: Final Answer:
The total number of niobium atoms present is $2.0 \times 10^{22}$, which corresponds to option (D).