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Silver crystallises in a face centered cubic unit cell . The density of Ag is $10.5 g cm^{-3}$. Calculate the edge length of the unit cell .

$(a)\;400pm\qquad(b)\;405pm\qquad(c)\;409pm\qquad(d)\;410pm$

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For face-centered cubic unit , Z = 4,
We know that , $V = \large\frac{Z\times M}{N_0\times d}$
$= \large\frac{4\times 108}{(6.023\times10^{23})\times10.5}$
$ = 6.83\times10^{-23}$
$= 68.3\times10^{-24}$
Let a be the edge length of the unit cell .
So, $V = a^3$
or $a^3 = 68.3\times10^{-24}$
$a = (68.3\times10^{-24})^{\large\frac{1}{3}}$cm
$ = 4.09\times10^{-8}$cm
= 409 pm
answered Apr 8, 2014 by sharmaaparna1
 

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