Question

$NaCl$ crystallizes in a FCC lattice. The edge length of unit cell is nearly $0.5nm$. Calculate the density of sodium chloride
Take $N_A=6\times {10}^{23}$
and
$\text{Molecular mass of}NaCl=58.5gmo{l}^{-1}$

• A. $0.33gc{m}^{-3}$
  
• B. $3.12gc{m}^{-3}$
• C. $1.55gc{m}^{-3}$
• D. $6.24gc{m}^{-3}$

Answer 1

The correct option is B $3.12gc{m}^{-3}$

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A\times a^3}$
where,
$Z=\text{No. of atoms in a unit cell}=4\left(FCC\right)M=\text{Molecular mass}{N}_A=\text{Avagadro number}$
$a=\text{Edge length of the unit cell}$

$\rho =\frac{Z\times M}{a^3\times N_A}=\frac{4\times 58.5}{(5\times {10}^{-8}{)}^3\times 6\times {10}^{23}}$

$\rho =3.12gc{m}^{-3}$