Single Choice

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:

A42.5 g/cm$$^{3}$$
B0.425 g/cm$$^{3}$$
C8.5 g/cm$$^{3}$$
Correct Answer
D4.25 g/cm$$^{3}$$

Solution

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

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

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

Upon substituting the values in the above density equation :

Density $$=\dfrac {2\times 213}{\left ( 436.6 \times 10^{-10} \right )^3 \times 6.022 \times 10^{23}} = 8.5 g/ cm^3$$

So the correct option is C.

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