Question

An element with BCC structure has edge length of 500 pm. If it's density is 4 g cm$^{-3}$, find atomic mass of the element?

• A. 150 g mol$^{-1}$
• B. 100 g mol$^{-1}$
• C. 125 g mol$^{-1}$
• D. 250 g mol$^{-1}$

Hint:

To make calculations lightning fast under exam conditions, notice that $4 \times 125 = 500$. Then, $500 \times 10^{-24} \times 10^{23} = 50$. Finally, $\frac{50}{2} = 25$, and $25 \times 6 = 150$. Keeping powers of ten organized makes physical chemistry arithmetic highly manageable!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem provides the crystal lattice type (BCC), the edge length of the unit cell ($a = 500\text{ pm}$), and the density of the crystal ($\rho = 4\text{ g cm}^{-3}$). We need to find the molar atomic mass ($M$) of the element.

Step 2: Key Formula or Approach:
The density of a crystalline solid unit cell is governed by the expression: $$\rho = \frac{Z \times M}{a^3 \times N_A}$$ Rearranging the formula to isolate the atomic mass ($M$): $$M = \frac{\rho \times a^3 \times N_A}{Z}$$ where: $\rho = \text{density} = 4\text{ g cm}^{-3}$ $Z = \text{number of atoms per unit cell for BCC} = 2$ $a = \text{edge length in cm} = 500\text{ pm} = 500 \times 10^{-10}\text{ cm} = 5 \times 10^{-8}\text{ cm}$ $N_A = \text{Avogadro's number} \approx 6.022 \times 10^{23}\text{ mol}^{-1}$

Step 3: Detailed Explanation:
1. First, calculate the volume of the cubic unit cell ($a^3$): $$a^3 = (5 \times 10^{-8}\text{ cm})^3 = 125 \times 10^{-24}\text{ cm}^3$$ 2. Substitute all values into the rearranged atomic mass expression: $$M = \frac{4\text{ g cm}^{-3} \times (125 \times 10^{-24}\text{ cm}^3) \times (6.022 \times 10^{23}\text{ mol}^{-1})}{2}$$ 3. Perform the mathematical calculations step-by-step: $$M = \frac{500 \times 10^{-24} \times 6.022 \times 10^{23}}{2}$$ $$M = \frac{500 \times 6.022 \times 10^{-1}}{2}$$ $$M = \frac{50 \times 6.022}{2} = 25 \times 6.022 = 150.55\text{ g mol}^{-1}$$ Rounding to the nearest whole integer gives approximately $150\text{ g mol}^{-1}$.

Step 4: Final Answer:
The atomic mass of the element is $150\text{ g mol}^{-1}$, matching option (A).