Q.An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm$^3$. How many atoms are present in 208 g of the element?

$m_{\text{cell}} = d\,a^3 = 7.2\times(2.88\times10^{-8})^3 = 1.72\times10^{-22}\text{ g}$

$N = \frac{208}{8.60\times10^{-23}} \approx 2.42\times10^{24}$

$a = 288\text{ pm} = 2.88\times10^{-8}\text{ cm}$

$a^3 = (2.88\times10^{-8})^3 = 2.389\times10^{-23}\text{ cm}^3$

$m_{\text{cell}} = d\times a^3 = 7.2\times 2.389\times10^{-23} = 1.720\times10^{-22}\text{ g}$

$m_{\text{atom}} = \frac{m_{\text{cell}}}{2} = \frac{1.720\times10^{-22}}{2} = 8.60\times10^{-23}\text{ g}$

$N = \frac{208}{8.60\times10^{-23}} = 2.42\times10^{24}\text{ atoms}$