Question

$CsBr$ crystallises in a body centred cubic lattice. The unit cell length is $436.6pm$ . Given that the atomic mass of $Cs=133$ and that of $Br=80$) amu and Avogadro number being $6.02\times {10}^{23}$$mo{l}^{-1}$, the density of $CsBr$ is:

• A. $8.25g/c{m}^3$
• B. $4.25g/c{m}^3$
• C. $42.5g/c{m}^3$
• D. $0.425g/c{m}^3$

Answer 1

Answer: (B)

$4.25g/c{m}^3$

According to the problem:-

Density of $C_{S}Br=\frac{2\times M}{a^3\times No}$

Given that:

Number of atoms in the $b.c.c$ unit cell $\left(Z\right)$=2

Molar mass of $CsBr\left(M\right)$=$133+80$=$213$

Edge length of unit cell $\left(a\right)$=$436.6pm$

=$436.6\times {10}^{-10}cm$

Therefore,

Density=$\frac{2\times 213}{\left(436.6\times {10}^{-10}\right)\times 6.02\times {10}^{23}}$

=$8.50g/c{m}^3$

For unit cell=$\frac{8.50}{2}=4.25g/c{m}^3$