JEE Advanced 2025 — Paper 2 — CHEM-III solution

JEE Advanced2025Paper 2CHEM-III

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 = $6 \times 1 0^{23}$ mol $^{- 1}$

Your answer

Solution

A ccp (fcc) unit cell contains $Z = 4$ atoms. With edge length $a = 400$ pm $= 400 \times 1 0^{- 10}$ cm $= 4 \times 1 0^{- 8}$ cm and $M = 105.6$ g mol $^{- 1}$ :

$ρ = \frac{Z M}{a ^{3} N _{A}} = \frac{4 \times 105.6}{( 4 \times 1 0 ^{- 8} ) ^{3} \times 6 \times 1 0 ^{23}}$

$a^{3} = 64 \times 1 0^{- 24}$ cm $^{3}$ , so denominator $= 64 \times 1 0^{- 24} \times 6 \times 1 0^{23} = 38.4$ .

$ρ = \frac{422.4}{38.4} = 11.00$ g cm $^{- 3}$ .

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