Question -
Consider an element ‘Argo’ which forms ccp lattice. Its X-ray studies show that the edge length of its unit cell is 4000 nm. Also the density of Argo is found to be 10g/cm3. Which of the following elements has the closest atomic mass to that of Argo.
Options:
A .  Copper (atomic mass= 63 g/mol)
B .  Silver (atomic mass = 108 g/mol)
C .  Molybdenum (atomic mass = 95 g/mol)
D .  Gold (atomic mass = 196 g/mol)
Answer: Option C : C The number of Argo atoms per unit cell(z) = 4 Density (d) = 10g/cm3 Edge length of unit cell (a) = 4000 nm = 4×10−8cm We know that, d=z×Ma3×NA=>M=d×a3×NAz=10×(4×10−8)3×6.023×10234=96.368 g/mol This value is closest to the atomic mass of molybdenum.
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