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An element (atomic mass =60) having face-centred cubic unit cell has a density of $6.23gcm^{-3}$.What is edge length of unit cell ?

$\begin{array}{1 1}400pm\\300pm\\200pm\\100pm\end{array} $

1 Answer

$a^3=\large\frac{z\times M}{N_A\times d}$
$\;\;\;\;=\large\frac{4\times 60}{6.023\times 10^{23}\times 6.23}$
$\Rightarrow a^3=64\times 10^{-24}cm^3$
$\Rightarrow a=(4\times 10^{-8}\times 10^{10})$pm=400pm
answered Aug 4, 2014 by sreemathi.v
 

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