Question

The element chromium exists in bcc lattice with unit cell edge $2.88\times {10}^{-10}m$. The density of chromium is $7.2\times {10}^{3}kg{m}^{-3}$. How many atoms does $52\times {10}^{-3}kg$ of chromium contain?

Answer 1

Let $M_A$ be molar mass of Chromium

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$Z=2$ , $a=2.88\mathring{A}$ , $N_A=6.023\times {10}^{23}$ , $\rho =7.2g/c{m}^3$

$7.2=\frac{2\times M_A}{6.022\times {10}^{23}\times 23.88\times {10}^{-24}}$

$M_A=52$ $amu$

So, no. of moles of Chromium= $\frac{52\times {10}^{-3}}{52}={10}^{-3}$ moles

No. of atoms= $6.023\times {10}^{23}\times {10}^{-3}=6.023\times {10}^{20}$ atoms