Question

The density of solid Ag is 10.5 g/cm3. How many atoms are present per cubic centimeter of Ag?

Answer 1

Answer:

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

Edge length = $4.1\times 10^{-8}\ cm$

Explanation:

(a)

Given that:-

The density of the solid Ag = 10.5 g/cm³

Molar mass of silver = 107.8682 g/mol

So, Moles present per cm³ of Ag = $\frac{10.5\ g/cm^3}{107.8682\ g/mol}$=0.0973 mol/cm³

Also, 1 mole = $6.023\times 10^{23}$ atoms.

So,

Atoms present per cm³ of Ag = $0.0973\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=5.8\times 10^{22}\ atoms/cm^3$

Thus, answer = $5.8\times 10^{22}\ atoms/cm^3$

In FCC, the number of atoms  in the unit cell = 4 unit cells

So,

Unit cells which are present per cubic centimeter of Ag = $\frac{5.8\times 10^{22}\ atoms/cm^3}{4}=1.45\times 10^{22}\ unit\ cells /cm^3$

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

(b)

The reciprocal of the unit cell/cm³ is the volume of the unit cell.

So, $Volume=\frac{1}{1.45\times 10^{22}\ unit\ cells /cm^3}=68.9\times 10^{-24}\ cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

(c)

Also, Volume = ${(Edge\ length)}^3$

Thus, edge length = ${Volume}^{\frac{1}{3}}$ = $\left(68.9\times \:\:10^{-24}\right)^{\frac{1}{3}}\ cm=4.1\times 10^{-8}\ cm$

Edge length = $4.1\times 10^{-8}\ cm$