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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
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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 = 400×10−10 cm d=2×12064×10−24×6×1023 =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=4r⇒r=√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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