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 $6.02\times {10}^{23}mo{l}^{-1}$ , the density of $CsBr$ is:
A.$42.5g/c{m}^3$
B.$0.425g/c{m}^3$
C.$8.5g/c{m}^3$
D.$4.25g/c{m}^3$

Answer 1

Hint:
Lattice structures are topologically ordered, three-dimensional open-celled structures composed of one or more repeating unit cells [2,3]. These cells are defined by the dimensions and connectivity of their constituent strut elements, which are connected at specific nodes.This question can be solved by using the formula $\rho ={\displaystyle \frac{Z\times M}{a^3\times N_A}}$ where $Z,M,a,N_A,\rho$ are number of atoms, Mass of consisting atom/molecule, edge length/unit cell length, Avogadro number and density respectively.
Value of $Z$ for BCC is 2.

Complete step by step answer:
Let’s deduce the expression $\rho ={\displaystyle \frac{Z\times M}{a^3\times N_A}}$ from the very beginning.
We know density of unit cell =mass of unit cell/volume of unit cell
Mass of a unit cell = number of atoms in unit cell $\times$ mass of single atom = $Z\times m---(1)$ where, $Z,m$ are number of atoms and mass of single atom/molecule in unit cell. We can find the mass of a single atom/molecular by dividing atomic mass $M$ by Avogadro’s Number $N_A$.
So, $m={\displaystyle \frac{M}{N_A}}---(2)$
From equation (1) & (2)
Mass of unit cell $={\displaystyle \frac{Z\times M}{N_A}}---(3)$
Now we can write density of unit cell $\rho ={\displaystyle \frac{Z\times M}{V\times N_A}}---(4)$ where $V$ is volume of unit cell. As we also know that the volume of a cube is cube of its edge length, consider an edge length $a$ , so volume $V=a^3$.
So, equation (4) can be rewritten as $\rho ={\displaystyle \frac{Z\times M}{a^3\times N_A}}---(5)$ .

Now, Let’s calculate the number of atoms in the unit cell that is BCC.
Atoms at the corners are 8 and contribution of each corner atom is 1/8 and atom at centre is one and it is not shared with any other cell.
So, $Z=8\times {\displaystyle \frac{1}{8}}+1=2$
As per question $a=436.6pm=436.6\times {10}^{-10}cm,M=133+80=213g/mol$
Putting all values in equation (5) we get,
 $\rho ={\displaystyle \frac{2\times 213g/mol}{{(436.6\times {10}^{-10}cm)}^3\times 6.02\times {10}^{23}mo{l}^{-1}}}$
 $\Rightarrow \rho =8.5g/c{m}^3$


Note:
While deducing the formula $\rho ={\displaystyle \frac{Z\times M}{a^3\times N_A}}$ we used $m$ and $M$ , both are different. $m$ is mass of single atom $M$ is atomic mass in amu. So, when we take $M$ atomic mass in grams, according to concept of moles, we get mass of 1 mole of atoms that are equal in Avogadro’s number $=6.02\times {10}^{23}mo{l}^{-1}$ . So, in order to find the mass of a single atom $m={\displaystyle \frac{M}{N_A}}$ .