Q12 / 150 · MHT CET 2026 (11 April Shift 2) · Chemistry
A compound forms bcc unit cell with edge length $400$ pm. Determine the molar mass of the compound if the density of compound is $3.5\ \text{g cm}^{-3}$ [$N_A = 6.022 \times 10^{23}$]
- A$65.2\ \text{g mol}^{-1}$
- Correct answerB$67.4\ \text{g mol}^{-1}$
- C$69.8\ \text{g mol}^{-1}$
- D$71.6\ \text{g mol}^{-1}$
Solution
The formula for the density of a unit cell is given by:
$d = \dfrac{Z \times M}{N_A \times a^3}$
For a bcc unit cell, the number of atoms per unit cell is $Z = 2$.
Given:
$d = 3.5\ \text{g cm}^{-3}$
$a = 400\ \text{pm} = 4 \times 10^{-8}\ \text{cm}$
$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$
Rearranging the formula to solve for molar mass $M$:
$M = \dfrac{d \times N_A \times a^3}{Z}$
Substituting the given values:
$M = \dfrac{3.5 \times 6.022 \times 10^{23} \times (4 \times 10^{-8})^3}{2}$
$M = \dfrac{3.5 \times 6.022 \times 10^{23} \times 64 \times 10^{-24}}{2}$
$M = \dfrac{3.5 \times 6.022 \times 6.4}{2}$
$M = 67.4464\ \text{g mol}^{-1}$
Rounding to one decimal place, $M \approx 67.4\ \text{g mol}^{-1}$.
Answer: $67.4\ \text{g mol}^{-1}$
$d = \dfrac{Z \times M}{N_A \times a^3}$
For a bcc unit cell, the number of atoms per unit cell is $Z = 2$.
Given:
$d = 3.5\ \text{g cm}^{-3}$
$a = 400\ \text{pm} = 4 \times 10^{-8}\ \text{cm}$
$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$
Rearranging the formula to solve for molar mass $M$:
$M = \dfrac{d \times N_A \times a^3}{Z}$
Substituting the given values:
$M = \dfrac{3.5 \times 6.022 \times 10^{23} \times (4 \times 10^{-8})^3}{2}$
$M = \dfrac{3.5 \times 6.022 \times 10^{23} \times 64 \times 10^{-24}}{2}$
$M = \dfrac{3.5 \times 6.022 \times 6.4}{2}$
$M = 67.4464\ \text{g mol}^{-1}$
Rounding to one decimal place, $M \approx 67.4\ \text{g mol}^{-1}$.
Answer: $67.4\ \text{g mol}^{-1}$
Hence, the correct answer is option: 'B'.