Question

A metal crystallises in BCC lattice with unit cell edge length of 300 pm and density 6.15 g cm^-3. The molar mass of the metal is

• A. 50 g mol^-1
• B. 60 g mol^-1
• C. 40 g mol^-1
• D. 70 g mol^-1

Answer 1

The Correct Option is A

The molar mass of the metal can be calculated using the formula for density:
\[ \text{Density} = \frac{Z \times M}{N_A \times V} \]
Where:
- \( Z \) is the number of atoms per unit cell (for BCC, \( Z = 2 \)),
- \( M \) is the molar mass of the metal,
- \( N_A \) is Avogadro's number (\(6.022 \times 10^{23}\) mol\(^{-1}\)),
- \( V \) is the volume of the unit cell.

Given:
The edge length \( a = 300 \, \text{pm} \), so the volume \( V \) is calculated as:
\[ V = a^3 = (300 \times 10^{-12} \, \text{m})^3 = 3 \times 10^{-29} \, \text{m}^3 \]
- Convert \( V \) to cm\(^3\):
\[ V = 3 \times 10^{-23} \, \text{cm}^3 \]

Now, using the formula to calculate molar mass \( M \):

\[ M = \frac{\text{Density} \times N_A \times V}{Z} \]
Substituting the given values:
\[ M = \frac{6.15 \times 6.022 \times 10^{23} \times 3 \times 10^{-23}}{2} = 50 \, \text{g/mol} \]

So, the correct answer is (A): 50 g/mol