Question

An element with a molar mass of $2.7\times {10}^{-2}$ $\mathrm{kg}{\mathrm{mol}}^{-1}$ forms a cubic unit cell with an edge length of $405pm$. If its density is $2.7\times {10}^3$ $\mathrm{kg}{\mathrm{m}}^{-3}$, the radius of the element is approximately ______ ${10}^{-12}\mathrm{m}$ (to the nearest integer).

Answer 1

The radius of the element can be calculated as follows:

Step 1:

Density$=$$\frac{ZM}{N_A{a}^3}$ $\mathrm{Z}$$=$ The number of atoms present in the unit cell.

$\mathrm{M}$ $=$Molar mass

$=2.7\times {10}^{-2}$$kg.mo{l}^{-1}$

$N_a=6.023\times {10}^{23}mo{l}^{-1}$

$\mathrm{a}$ $=$edge length of unit cell

$=$$405pm$

$=405\times {10}^{-12}m$

Density $=$$2.7\times {10}^{3}kg{m}^{-3}$

Step 2:

So, Density $=$$\frac{Z\times 2.7\times {10}^{-2}}{6.023\times {10}^{23}\times {\left(405\times {10}^{-12}\right)}^3}$

$Z=6.023\times 405\times 405\times 405\times {10}^{-8}$

$=4$

So we can say, this is a face-centered cubic system.

Step 3:

For face centered cubic system:

$4\mathrm{r}=\mathrm{a}\times \sqrt{2}$

$$\begin{gathered} 2\mathrm{r}\times \sqrt{2}=405\mathrm{pm} \\ 2\mathrm{r}\times \sqrt{2}=405\times {10}^{-12}\mathrm{m} \\ \mathrm{r}=143\times {10}^{-12}\mathrm{m} \end{gathered}$$

Hence, the radius of the element is approximately $143$ $\times$${10}^{-12}\mathrm{m}$.