JEE Advanced 2025 · Paper 2 · Chemistry Q9

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ChemistryNumerical+4Question 41 of 48

The density (in $\mathrm{g\,cm^{-3}}$ ) 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 $= 6\times10^{23}\ \mathrm{mol^{-1}}$

Your answer

Answer and worked solution

Official answer: 10.85

A ccp (fcc) cell has $Z=4$ atoms and edge $a=400$ pm $=4\times10^{-8}$ cm.
$\rho=\frac{ZM}{N_Aa^3}=\frac{4\times105.6}{6\times10^{23}\times64\times10^{-24}}=\frac{422.4}{38.4}=11.0\ \text{g cm}^{-3}.$

Topics:Density of crystals Crystal lattices and unit cells

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JEE Advanced 2025 · Paper 2, Chemistry question 9