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

1. Identify the given information:

- Crystal lattice type: BCC (Body-Centered Cubic)
- Unit cell edge length ($a$) = 300 pm
- Density ($\rho$) = 6.15 g cm^-3
- We need to find the Molar Mass ($M$).
- We know Avogadro's number ($N_A$) ≈ 6.022 x 10²³ mol^-1.

2. Determine the number of atoms per unit cell (Z) for BCC:

- In a BCC lattice, there is 1 atom at the body center and 1/8 of an atom at each of the 8 corners.
- $Z = 1 (\text{center}) + 8 \times (1/8) (\text{corners}) = 1 + 1 = 2$ atoms per unit cell.

3. Convert the unit cell edge length ($a$) to cm:

- 1 pm = 10^-10 cm
- $a = 300 \, \text{pm} = 300 \times 10^{-10} \, \text{cm} = 3 \times 10^2 \times 10^{-10} \, \text{cm} = 3 \times 10^{-8} \, \text{cm}$

4. Calculate the volume of the unit cell ($V$):

- For a cubic cell, $V = a^3$
- $V = (3 \times 10^{-8} \, \text{cm})^3 = 3^3 \times (10^{-8})^3 \, \text{cm}^3 = 27 \times 10^{-24} \, \text{cm}^3$

5. Use the density formula for crystals:

- Density ($\rho$) = $\frac{Z \times M}{N_A \times V}$
- Where:
    $\rho$ = density
    $Z$ = number of atoms per unit cell
    $M$ = molar mass (in g/mol)
    $N_A$ = Avogadro's number
    $V$ = volume of the unit cell

6. Rearrange the formula to solve for Molar Mass ($M$):

- $M = \frac{\rho \times N_A \times V}{Z}$

7. Plug in the values and calculate $M$:

- \(M = \frac{(6.15 \, \text{g cm}^{-3}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \times (27 \times 10^{-24} \, \text{cm}^3)}{2}\)
- $M = \frac{(6.15) \times (6.022) \times (27) \times 10^{23 - 24}}{2} \, \text{g mol}^{-1}$
- $M = \frac{(6.15) \times (6.022) \times (27) \times 10^{-1}}{2} \, \text{g mol}^{-1}$
- $M = \frac{(6.15) \times (6.022) \times 2.7}{2} \, \text{g mol}^{-1}$
- $M \approx \frac{(37.03) \times 2.7}{2} \, \text{g mol}^{-1}$
- $M \approx \frac{99.98}{2} \, \text{g mol}^{-1}$
- $M \approx 49.99 \, \text{g mol}^{-1}$

Therefore, the molar mass of the metal is 50 g mol^-1.

Final Answer: The final answer is $50 \, \text{g mol}^{-1}$.

Answer 2

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