Question

An element germanium crystallises in bcc type crystal structure with edge of unit cell $288pm$ and the density of the element is $7.2gc{m}^{-3}$. Calculate the number of atoms present in $52g$ of the crystalline element. Also calculate atomic mass of the element.

Answer 1

Given that,

Mass of atoms, $w=52g$

Density, $d=7.2g.{cm}^{-3}$

Edge length, $a=288\times {10}^{-10}cm$

$N_A=6.02\times {10}^{23}$

$Z=2$ for $BCC$

Now,

$d=\frac{Z\times M}{N_0\times a^3}$

$M=\frac{d\times a^3\times N_A}{Z}=\frac{7.25\times (2.88\times {10}^{-10})\times 6.02\times {10}^{23}}{2}$

$M=51.77g/mol$

Hence the atomic mass of the element is $51.77u$.

$\therefore$ Number of atoms, $N=\frac{Mass}{Molarmass}\times N_A=\frac{52}{51.77}\times 6.02\times {10}^{23}=6.05\times {10}^{23}atoms$