Question

A metal has $\mathrm{BCC}$ structure with edge length of unit cell $400\ \mathrm{pm}$. Density of metal is $4\ \mathrm{g\ cm^{-3}}$. What is molar mass of metal?

• A. $40\ \mathrm{g\ mol^{-1}}$
• B. $27\ \mathrm{g\ mol^{-1}}$
• C. $92\ \mathrm{g\ mol^{-1}}$
• D. $77\ \mathrm{g\ mol^{-1}}$

Hint:

To streamline calculations during exams, remember that $10^{-24} \times 10^{23} = 0.1$. Then group variables simply: $M = \frac{\text{Density} \times a^3 \times 0.6}{2}$. This simplifies the arithmetic step down to a few basic products!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a metallic crystalline structure that crystallizes in a body-centered cubic ($\mathrm{BCC}$) lattice setup. The cell parameters given are the edge length ($a = 400\ \mathrm{pm}$) and the density ($\rho = 4\ \mathrm{g\ cm^{-3}}$). We need to calculate the molar mass ($M$) of this metal.

Step 2: Key Formula or Approach:
The standard equation linking density and cell dimensions is: $$\rho = \frac{Z \times M}{a^3 \times N_A} \implies M = \frac{\rho \times a^3 \times N_A}{Z}$$ Where:

• $Z = 2$ (for a body-centered cubic unit cell).

• $N_A \approx 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$ (Avogadro's constant).

• $a = 400\ \mathrm{pm} = 400 \times 10^{-10}\ \mathrm{cm} = 4 \times 10^{-8}\ \mathrm{cm}$.

Step 3: Detailed Explanation:
Let's calculate the volume of the unit cell ($a^3$): $$a^3 = (4 \times 10^{-8}\ \mathrm{cm})^3 = 64 \times 10^{-24}\ \mathrm{cm^3}$$ Substitute all values into the rearranged molar mass equation: $$M = \frac{4\ \mathrm{g\ cm^{-3}} \times (64 \times 10^{-24}\ \mathrm{cm^3}) \times (6.022 \times 10^{23}\ \mathrm{mol^{-1}})}{2}$$ Simplify the constants: $$M = \frac{4 \times 64 \times 6.022 \times 10^{-1}}{2}$$ $$M = 2 \times 64 \times 0.6022$$ $$M = 128 \times 0.6022 \approx 77.08\ \mathrm{g\ mol^{-1}}$$ This closely matches $77\ \mathrm{g\ mol^{-1}}$.

Step 4: Final Answer:
The molar mass of the metal is $77\ \mathrm{g\ mol^{-1}}$, which corresponds to option (D).