Question

An element occurs in the body-centred cubic structure with an edge length of 288 pm. The density of the element is 7.2 g cm$^{-3}$. The number of atoms present in 208 g of the element is nearly

• A. $24.2 \times 10^{23}$
• B. $12.1 \times 10^{23}$
• C. $24.2 \times 10^{24}$
• D. $36.3 \times 10^{23}$

Hint:

BCC: Z=2. Density = $\frac{ZM}{N_Aa^3}$.

Answer 1

The Correct Option is A

For a body-centred cubic (BCC) structure, the number of atoms per unit cell is $Z = 2$. The relationship between density ($\rho$), molar mass ($M$), Avogadro's number ($N_A$), edge length ($a$), and $Z$ is: $$ \rho = \frac{ZM}{N_Aa^3} $$ We can rearrange this to find the molar mass: $$ M = \frac{\rho N_A a^3}{Z} $$ Plugging in the values: $$ M = \frac{(7.2 \, \text{g cm}^{-3})(6.022 \times 10^{23} \, \text{atoms mol}^{-1})(288 \times 10^{-10} \, \text{cm})^3}{2} \approx 52 \, \text{g/mol} $$ Number of moles in 208 g = $\frac{208 \, \text{g}}{52 \, \text{g/mol}} = 4 \, \text{mol}$ Number of atoms = $4 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 24.088 \times 10^{23} \, \text{atoms} \approx 24.2 \times 10^{23}$