Chemistry · Solid State

JEE Advanced 2025 — Paper 2 — Question 38

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 $\qquad$ .

Use: Atomic mass of metal $=105.6 \mathrm{amu}$ and Avogadro's constant $=6 \times 10^{23} \mathrm{~mol}^{-1}$

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

( $\mathrm{Z}=4, \mathrm{a}=400 \mathrm{pm}, \mathrm{M}=105.6 \mathrm{~g} / \mathrm{mol}$ )

Density $=\frac{Z \times M}{a^{3} \times N_{A}}$

$=\frac{4 \times 105.6}{\left(400 \times 10^{-10}\right)^{3} \times 6 \times 10^{23}}=11.00 \mathrm{~g} / \mathrm{cm}^{3}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Chemistry
Chapter: Solid State
Topic: Types of Crystal Packing: SC, BCC, FCC, Hexagonal