Question

Silver crystallizes in the fcc lattice. If the edge length of the cell is $4.07*{10}^{-8}$ cm and the density is $10.5\mathrm{g}{\mathrm{cm}}^{-3}$. Calculate the atomic mass of metal?

Answer 1

Step 1:Given data:

Silver crystallizing in the fcc lattice

Hence, the number of atoms per unit cell $\left(\mathrm{Z}\right)$ for Fcc lattice $=4$

The edge length of the cell is given as $\left(\mathrm{a}\right)=4.07*{10}^{-8}\mathrm{cm}$

The density of the unit cell is given as = $10.5\mathrm{g}{\mathrm{cm}}^{-3}$

Step 2:Formula for density of the unit cell:

The density of a unit cell can be given as:

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

Where,

$\mathrm{Z}$ =The number of atoms per unit cell

$\mathrm{a}=$ The edge length of the cell

$\mathrm{D}=$The density of the unit cell

${\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}$=Avogadro’s number
Step 3: Calculation of the atomic mass of Silver:

By substituting the values in the density formula, we can get:

$$\begin{gathered} 10.5=\frac{4\times \mathrm{M}}{(6.023\times {10}^{23})\times {(4.07\times {10}^{-8})}^3} \\ \Rightarrow \mathrm{M}=107.9\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Hence, the atomic mass of the Silver metal is calculated to be $107.9\mathrm{g}{\mathrm{mol}}^{-1}$