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Cu metal crystallises in face centered cubic lattice with cell edge , a = 361.6 pm . What is the density of Cu crystal ? (Atomic mass of copper = 63.5 amu , $N_A = 6.023\times10^{23}$

$(a)\;89.4g\;cm^{-3}\qquad(b)\;8.94g\;cm^{-3}\qquad(c)\;894g\;cm^{-3}\qquad(d)\;0.894g\;cm^{-3}$

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We know that
$Z = \large\frac{a^3\times d\times N_A}{M}$
$\therefore d = \large\frac{ZM}{a^3N_A}$
Z = 4 for fcc unit cell ;
$M = 63.5 g\; mol^{-1}$
$a = 361.6\times 10^{-10}cm$
$N_A = 6.023\times10^{23}$
Putting these values in (i) we get
$d = \large\frac{4\times63.5}{(361.6\times10^{-10})^3\times 6.023\times10^{23}}$
$=8.94 g\;cm^{-3}$
Hence answer is (B)
answered Apr 9, 2014 by sharmaaparna1
 
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