Question

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

Answer 1

In bcc struture per unit cell Z = 2

a = 288 pm =$288\times {10}^{-}8cm$

Volume(V) = $a^3$ = $2.39\times {10}^{-}23c{m}^3$

Density (d)= $7.2g/c{m}^3$

$N_A$ = Avogadro constant = $6.022\times {10}^{2}3$

Molecular mass (M) = /

we know $M=\frac{d\times N_A\times a^3}{Z}$

on substituting values

M = $\frac{7.2\times (6.022\times {10}^{2}3)\times (2.39\times {10}^{-}23)}{2}$

=$51.8g$

or $52g$

so 52 gram of this element have one mole so it contains $6.022\times {10}^{23}$ atoms.

And atomic mass = $52/6.022\times {10}^{23}=8.633\times {10}^{-23}$