Question

The density of silver having an atomic mass of 107.8 g mol^{- 1} is 10.8 g cm^-3. If the edge length of cubic unit cell is 4.05 × 10^{- 8}
 cm, find the number of silver atoms in the unit cell.
 ( N_A = 6.022 × 10²³, 1 Å = 10^-8 cm)

Answer 1

Given:
Density (d) = 10.8 g cm^-3
Edge length (a) = 4.05 x 10^{- 8} cm
Molar mass = 107.8 g mol^-1
Avogadro's number (N_A) = 6.022 x 10²³

To find:
Number of atoms in the unit cell

Formula:
a. Mass of one atom = `"Atomic mass"/"Avogadro number"`
b. Volume of unit cell = a³
c. Density = `"Mass of unit cell"/"Volume of unit cell"`

Calculation:
a) Mass of one Ag atom = `"Atomic mass of Ag"/"Avogadro number"`
Avogadro number
= `107.8/(6.022 xx 10^23)`

= 1.79 x 10^-22 g

b) Volume of unit cell = a³
= ( 4.05 x 10^-8 )³
= 6.64 x 10^-23 cm³

c)
Density (d) = `"Mass of unit cell"/"Volume of unit cell"`

= `"Number of atoms in unit cell x Mass of one atom"/"Volume of unit cell"`

10.8 = `("Number of atoms in unit cell" xx 1.79 xx 10^-22)/(6.64 xx 10^-23)`

Number of atoms in unit cell = `( 10.8 xx 6.64 xx 10^-23)/( 1.79 xx 10^-22)`

= 40.06 x 10^-1 = 4.0 ≈ 4

∴ The number of atoms in the unit cell of silver is 4.