Question

An element crystallises in face-centred cubic lattice with density as $5g/c{m}^3$ and edge length of the side of unit cell is 400 pm.
Take $N_A=6\times {10}^{23}$ Then the molar mass of the element in $gmo{l}^{-1}$ is :

• A. 48.00
• B. 48
• C. 48.0

Answer 1

Answer: (A), (B), (C)

$Z=4$ in fcc lattice
$\rho =5g/c{m}^3$
$a=400pm=4\times {10}^{-8}cm$
$\therefore a^3=64\times {10}^{-24}c{m}^3$
$N_A=6\times {10}^{23}$
$\rho =\frac{Z\times M}{N_A\times a^3}$

$\therefore M=\frac{\rho N_A{a}^3}{Z}$
$M=\frac{5g/c{m}^3\times 6\times {10}^{23}{\text{mol}}^{-1}\times 64\times {10}^{-24}c{m}^3}{4}$

$M=48gmo{l}^{-1}$