Question

An element exists in bcc lattice with a cell edge of $300pm$. Calculate the molar mass if density of the unit cell is $7g//c{m}^3$.
Take $N_A=6\times {10}^{23}$

• A. $56.7gmo{l}^{-1}$
• B. $5.67gmo{l}^{-1}$
• C. $113.4gmo{l}^{-1}$
• D. $11.34gmo{l}^{-1}$

Answer 1

The correct option is A $56.7gmo{l}^{-1}$

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A(a^3\times {10}^{-30})}gc{m}^{-3}$
where,
$Z=\text{No. of atoms in a unit cell}=2\left(\text{for BCC}\right)M=\text{Molar mass}{N}_A=\text{Avagadro number}a=\text{Edge cell length in pm}$
$7=\frac{2\times M}{6\times {10}^{23}\times (300\times {10}^{-10}{)}^3}M=56.7gmo{l}^{-1}$