Question

A metal of density $7.08gc{m}^{-3}$ at $298K$ crystallizes in a body centered cubic lattice with lattice constant $2.9\stackrel{\circ }{A}$. The metal is (atomic mass is given in parenthesis) :

• A. $Cr\left(52.0\right)$
• B. $Zn\left(65.4\right)$
• C. $Sc\left(45.0\right)$
• D. $Rh\left(102.9\right)$

Answer 1

The correct option is A $Cr\left(52.0\right)$

A body centered cubic lattice has 2 atoms per unit cell $(Z=2)$.
The expression for density $\left(\rho \right)$ is
$\rho =\frac{M\times Z}{N_0\times a^3}$
Here, $M$ is the molecular weight of metal (in g/mol), Z is the number of atoms of metal per unit cell, $N_0$ is avogadro's number and ${}^{\prime }{a}^{\prime }$ is the lattice parameter (edge length).
$a=2.9A^0=2.9\times {10}^{-8}cm$
$7.08g/c{m}^3=\frac{M\times 2}{6.023\times {10}^{23}\times (2.9\times {10}^{-8}cm{)}^3}$
$M=52.0g/mol$ which is the molecular weight of $Cr$.