Question

The element chromium exists as a bcc lattice whose unit cell edge is $2.88\mathring{A}$. The density of chromium is $7.20$ g cm${}^{-3}$. How many atoms does $52.0$ g of chromium contain?

• A. $3.1\times {10}^{23}$ atoms mol${}^{-1}$
• B. $6.05\times {10}^{23}$ atoms mol${}^{-1}$
• C. $1.2\times {10}^{24}$ atoms mol${}^{-1}$
• D. $12\times {10}^{24}$ atoms mol${}^{-1}$

Answer 1

The correct option is B $6.05\times {10}^{23}$ atoms mol${}^{-1}$

As we know, density $=\frac{zM}{V{N}_o}$, where $n$ is the number of atoms per unit cell.

Here, $z=2$ for bcc lattice

$7.2=\frac{2\times 52}{(2.88\times {10}^{-10}{)}^3(N_o)}$

$⟹N_o=6.05\times {10}^{23}$ atoms mol${}^{-1}$

Hence, the correct option is $\text{B}$