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A metal (atomic mass =50) has a body-centred cubic lattice.The density of the metal is $5.91gcm^{-3}$.Find out the volume of the unit cell.$[N_A=6.023\times 10^{23}]$

$\begin{array}{1 1}2.809\times 10^{-23}cm^3\\1.809\times 10^{-23}cm^3\\0.809\times 10^{-23}cm^3\\4.809\times 10^{-23}cm^3\end{array} $

1 Answer

$N_A=6.023\times 10^{23}$
$d=\large\frac{z\times M}{N_A\times a^3}$
$a^3=\large\frac{z\times M}{N_A\times d}$
$\;\;\;\;=\large\frac{2\times 50}{5.91\times 6.023\times 10^{23}}$
$\;\;\;\;=2.809\times 10^{-23}cm^3$
answered Aug 4, 2014 by sreemathi.v

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