Question

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

• A. 42.5 g/cm${}^3$
• B. 0.425 g/cm${}^3$
• C. 8.5 g/cm${}^3$
• D. 4.25 g/cm${}^3$

Answer 1

Answer: (C)

8.5 g/cm${}^3$

Denisty $=\frac{Z\times M}{a^3\times N_A}$

Given $Z=$ two formula unit of CsBr in 1 cubic unit $=$ 2 and edge length, $a=436.6pm=436.6\times {10}^{-10}cm$.

Molecular weight of $CsBr=133+80=213$ g/mol.

Upon substituting the values in the above density equation :

Density $=\frac{2\times 213}{{(436.6\times {10}^{-10})}^3\times 6.022\times {10}^{23}}=8.5g/c{m}^3$

So the correct option is C.