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Question

# An element crystallises in a face-centered cubic lattice having density of 5 g/cm3 and edge length of the side of unit cell is 300 pm. Calculate mass of the element containing 3×1024 atoms. Take NA=6×1023

A
10.12 g
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B
145.66 g
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C
20.25 g
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D
101.25 g
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Solution

## The correct option is D 101.25 g Density of the unit cell, ρ=Z×MNA×a3 where, Z=No. of atoms in a unit cell=4 for FCCM=Molecular massNA=Avagadro number=6×1023 a=length of the unit cell =300pm=3×10−8cm ∴M=ρNAa3Z M=5×6×1023×27×10−244 M=20.25 g mol−1 Thus, 6×1023 atoms weigh 20.25 g 3×1024 atoms weigh 20.25×3×10246×1023=101.25 g

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