JEE Advanced 2025 Paper 2 · Q09 · Atomic Structure
The density (in $g/cm^3$) 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 amu and Avogadro's constant = $6 \times 10^{23}$ $mol^{-1}$.
Reveal answer + step-by-step solution
Correct answer:11
Solution
ccp -> $Z = 4$ atoms per unit cell. $\rho = Z \cdot M / (N_A \cdot a^3)$. $a = 400$ pm = $4 \times 10^{-8}$ cm; $a^3 = 64 \times 10^{-24}$ $cm^3 = 6.4 \times 10^{-23}$ $cm^3$. $\rho = (4 \times 105.6) / (6 \times 10^{23} \times 6.4 \times 10^{-23}) = 422.4 / 38.4 = 11$ $g/cm^3$.
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