Question

Copper crystallises into a fcc lattice with edge length 3.61 × 10 −8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm −3 .

Answer 1

Hey there!

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Answer :

$d = \frac{Z × M}{a^{3} × N_{A}}$

⇒ $d = \frac{4×63.5}{(3.61)^{3} × 10^{-24} × 6.022 × 10^{23}}$

⇒ $d = \frac{254.0 × 10}{47.045 × 6.022}$

⇒ $d = \frac{2540}{283.35}$

⇒ d = 8.96 g $cm^{-3}$

Answer 2

Given, edge length (a) = 3.61 × 10^-8 cm
Given lattice is fcc, therefore, no. of atoms per unit cell (z) = 4
We know, atomic mass of copper, M= 63.5 g/mol
Avogadro number, $N_A$ = 6.022 × 10²³ g/mol
To prove, density (d) = 8.92 g/cm³
we know that, density $\bf{\rho=\frac{zM}{a^3N_A}}$
= (4 × 63.5 )/{(3.61 × 10^-8)³ × 6.023 × 10²³}
= 254/(47.0458 × 10^-24 × 6.023 × 10²³)
= 2540/283.30
= 8.96 g/cm³

hence, it is clear that calculated density is in agreement with its measured value of 8.92 g/cm³