Question

$\mathrm{CsBr}$crystallises in a body-centred cubic lattice. The unit cell length is $436.6\mathrm{pm}.$ Given that the atomic mass of $\mathrm{Cs}\mathrm{is}133,\mathrm{Br}\mathrm{is}80\mathrm{amu}$ and Avogadro number is $6.02\times {10}^{23}{\mathrm{mol}}^{-1}$. What will be the density of $\mathrm{CsBr}$?

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

Answer 1

The correct option is A

$8.5\mathrm{g}{\mathrm{cm}}^{-3}$

Explanation for the correct option:

Option (A):$8.5\mathrm{g}{\mathrm{cm}}^{-3}$

Step 1:Given data:

The edge length of the unit cell is given as ,$\mathrm{a}=436.6\mathrm{pm}=436.6\times {10}^{-10}\mathrm{cm}.$

As the atomic mass of $\mathrm{Cs}\mathrm{is}133,\mathrm{Br}\mathrm{is}80\mathrm{amu}$

Molecular weight of $\mathrm{CsBr}=133+80=213\mathrm{g}{\mathrm{mol}}^{-1}$

Given $\mathrm{Z}$,as the number of unit cells of $\mathrm{CsBr}$in 1 cubic unit $=2$

Also given that, the Avogadro number is $6.02\times {10}^{23}{\mathrm{mol}}^{-1}$.

Step 2: Formula for calculating the density of unit cell:

$\mathrm{Denisty}=\frac{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{Z}\times \mathrm{M}}$​

Step 3: Calculating the density of unit cell:

Upon substituting the values in the above density equation we can calculate the density of the unit cell as:

$$\begin{gathered} \mathrm{Density}=\frac{{(436.6\times {10}^{-10})}^3\times 6.022\times {10}^{23}}{2\times 213} \\ =8.5\mathrm{g}{\mathrm{cm}}^{-3} \end{gathered}$$

Hence, the density of $\mathrm{CsBr}$ is calculated as $8.5\mathrm{g}{\mathrm{cm}}^{-3}$.