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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.
( NA = 6.022 × 1023, 1 Å = 10-8 cm)
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Solution
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 (NA) = 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.
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