Question

A metal (Atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is $5.96g/c{m}^3$. Find the volume of unit cell. $(N_0=6.023\times {10}^{23}atomsmo{l}^{-1})$

• A. $2.8\times {10}^{-29}c{m}^3$
• B. $2.8\times {10}^{-23}c{m}^3$
• C. $12.8\times {10}^{-30}c{m}^3$
• D. $14.6\times {10}^{-15}c{m}^3$

Answer 1

Answer: (B)

$2.8\times {10}^{-23}c{m}^3$

Density $\rho =\frac{Z\times M}{N_0\times a^3}$
$\rho =5.96g/c{m}^3=$ density
$Z=2atoms=$ numeber of atoms per unit cell.
$M=50g/mol=$ atomic weight of metal

$N_0=6.023\times {10}^{23}atoms/mol=$Avogadro's number

$a=$ edge length

Density $\rho =\frac{Z\times M}{N_0\times a^3}$

Density $5.96g/c{m}^3=\frac{2atoms\times 50g/mol}{6.023\times {10}^{23}atoms/mol\times (a{)}^3}$

$V=a^3=2.8\times {10}^{-23}c{m}^3$

$V$ is the volume of unit cell.