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 $10g/c{m}^3$. Which of the following elements has the closest atomic mass to that of Argo.

• 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 1

The correct option is C Molybdenum (atomic mass = 95 g/mol)

The number of Argo atoms per unit cell(z) = 4
Density (d) = $10g/c{m}^3$
Edge length of unit cell (a) = 4000 nm = $4\times {10}^{-8}$cm
We know that, $d=\frac{z\times M}{a^3\times N_A}=>M=\frac{d\times a^3\times N_A}{z}=\frac{10\times (4\times {10}^{-8}{)}^3\times 6.023\times {10}^{23}}{4}=96.368$ g/mol
This value is closest to the atomic mass of molybdenum.