Q.An element has bcc structure with a cell edge of 288 pm. The density of the element is 7.2 gcm$^{-3}$. How many atoms are present in 208g of the element.

Step 1. Convert the edge length: $a = 288$ pm $= 2.88\times10^{-8}$ cm, so $a^3 = (2.88\times10^{-8})^3 = 2.389\times10^{-23}\ \text{cm}^3$.

Step 2. Mass of one unit cell $= \rho \times a^3 = 7.2 \times 2.389\times10^{-23} = 1.720\times10^{-22}$ g.

Step 3. Since the structure is bcc, this mass corresponds to $n=2$ atoms, so mass of one atom $= \dfrac{1.720\times10^{-22}}{2} = 8.600\times10^{-23}$ g.

Step 4. Molar mass $M = (\text{mass of one atom})\times N_A = 8.600\times10^{-23}\times6.023\times10^{23} \approx 51.8\ \text{g mol}^{-1}$ (consistent with chromium, atomic mass ~52). …

Use rho = zM/(a^3 NA) to first find M from the given density and edge length, then convert 20 …

• Trying to find the atom count without first solving for M -- since M is not given directly, it must be extracted from the densi …