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Copper crystallises into a fcc lattice with edge length $3.6\times 10^{-8}$cm.Show that the calculated density is in agreement with its measured value of $8.92gcm^{-3}$.

1 Answer

$d=\large\frac{z\times M}{a^3\times N_A}$
No. of atoms in fcc lattice z=4
$d=\large\frac{4\times 63.5mol^{-1}}{(3.61\times 10^{-8}cm)^3\times (6.02\times 10^{23}mol^{-1})}$
Atomic mass of unit cell =$\large\frac{4\times 63.5}{6.02\times 10^{23}}$g
$\Rightarrow 4.22\times 10^{-22}$g
Density =$\large\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$
$\Rightarrow \large\frac{4.22\times 10^{-22}}{47.4\times 10^{-24}cm^2}$
$\Rightarrow 8.9gcm^{-3}$
Which is in close agreement with the measured value.
Density = Mass of unit cell/Volume of unit cell
$8.93 = (4 \times 63.5)/(a^3 \times 6.02 \times 10^{23}).$
$a^ 3 = 47.24 x 10^{-24 }$
$\qquad= 360\;pm$
answered Jul 30, 2014 by sreemathi.v
edited Oct 30 by meena.p
 

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