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Question

An element A (Atomic weight = 120) having bcc structure has unit edge length 400 pm. Identify the correct option(s). (Given: NA=6×1023)

A
The density of solid element is 6.25 g/cm3
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B
There are 6×1022 unit cells in 24 g of the solid element
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C
The atomic radius is about 1.732 A
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D
In 25 g of solid element, the volume occupied by atoms is nearly 2.72 cm3 only
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Solution

The correct options are
A The density of solid element is 6.25 g/cm3
B There are 6×1022 unit cells in 24 g of the solid element
C The atomic radius is about 1.732 A
D In 25 g of solid element, the volume occupied by atoms is nearly 2.72 cm3 only
(A) Density of the solid element
d=Z×Ma3×NA
Where,
Z is the number of atoms per unit cell
M is molar mass of the atom
a is the edge length of unit cell
NA is the Avogadro number
Given,
Z for BCC structure = 2
Molar mass = 120 g/mol
Edge length = 400×1010 cm

d=2×12064×1024×6×1023

=406.4=10.16 g/cm3=6.25 g/cm3

(B) Number of atoms per unit cell for BCC = 2
Number of unit cell =number of atoms2
Number of atoms =number of moles×NA

Number of moles =massmolar mass
Given mass = 24 g
number of moles =24120=0.2
Number of atoms =0.2×6×1023

Number of unit cell =12×10222
=6×1022

(C) The relationship between radius and edge length for BCC is given by
3a=4rr=34×400
=1.732×100 pm=1.732 A

(D) Given mass = 25 g and density = 6.25 g/cm3
Volume of the unit cell =massdensity=256.25
Fraction of the volume occupied by atoms in BCC = 0.68
hence,
Voccupied=0.68×256.25=2.72 cm3

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