JEE Advance - Chemistry (2025 - Paper 2 Online - No. 9)
The density (in $\mathrm{g} \mathrm{cm}^{-3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ___________.
Use: Atomic mass of metal $=105.6 \mathrm{amu}$ and Avogadro's constant $=6 \times 10^{23} \mathrm{~mol}^{-1}$
Explanation
• In a ccp (fcc) unit cell, number of atoms per cell, n = 4
• Edge length
$$a=400\text{ pm}=400\times10^{-12}\text{ m}=4.0\times10^{-8}\text{ cm}$$
• Volume of cell
$$V=a^3=(4.0\times10^{-8}\text{ cm})^3=6.4\times10^{-23}\text{ cm}^3$$
• Mass of one atom
$$m_{\rm atom}=\frac{M}{N_A}=\frac{105.6\text{ g/mol}}{6.0\times10^{23}\text{ mol}^{-1}} =1.76\times10^{-22}\text{ g}$$
• Mass of unit cell
$$m_{\rm cell}=n\;m_{\rm atom}=4\times1.76\times10^{-22}=7.04\times10^{-22}\text{ g}$$
• Density
$$\rho=\frac{m_{\rm cell}}{V} =\frac{7.04\times10^{-22}}{6.4\times10^{-23}} \approx11\;\text{g/cm}^3$$
Answer: 11 g·cm⁻³.
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