Question

An element crystallises in a face-centered cubic lattice having density of $5g/c{m}^3$ and edge length of the side of unit cell is $300pm$. Calculate mass of the element containing $3\times {10}^{24}$ atoms.
Take
$N_A=6\times {10}^{23}$

• A. $10.12g$
• B. $145.66g$
• C. $20.25g$
• D. $101.25g$

Answer 1

The correct option is D $101.25g$

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A\times a^3}$
where,
$Z=\text{No. of atoms in a unit cell}=4\text{for FCC}M=\text{Molecular mass}{N}_A=\text{Avagadro number}=6\times {10}^{23}$
$a=\text{length of the unit cell}=300pm=3\times {10}^{-8}cm$


$\therefore M=\frac{\rho N_A{a}^3}{Z}$
$M=\frac{5\times 6\times {10}^{23}\times 27\times {10}^{-24}}{4}$

$M=20.25gmo{l}^{-1}$

Thus,
$6\times {10}^{23}\text{atoms weigh}20.25g$
$3\times {10}^{24}\text{atoms weigh}\frac{20.25\times 3\times {10}^{24}}{6\times {10}^{23}}=101.25g$