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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. - Chemistry

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प्रश्न

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)

संख्यात्मक
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उत्तर

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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