Question

The density of $Ca{F}_2$ is $3.180gc{m}^{-3}$ and edge length, a is $0.55nm$. Then number of particles $\left(Z_{eff}\right)$ in a unit cell are nearly :
Given :
$\text{Molar mass of}Ca{F}_2=78gmo{l}^{-1}$

• A. 2
• B. 4
• C. 8
• D. 12

Answer 1

The correct option is B 4

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A{a}^3}$
where,
$Z=\text{No. of atoms in a unit cell}M=\text{Molar mass}{N}_A=\text{Avagadro number}$

$\text{Edge length, a}=0.55nma=0.55\times {10}^{-7}cm$
$\therefore Z_{eff}=\frac{\rho N_A{a}^3}{M}$
$=\frac{3.18gc{m}^{-3}\times 6.02\times {10}^{23}{\text{mol}}^{-1}\times (0.55\times {10}^{-7}cm{)}^3}{78g{\text{mol}}^{-1}}=4.08\approx 4$
$Z_{eff}=4$

Thus, the unit cell contains 4 $Ca{F}_2$ molecules.
Thus, $C{a}^{2+}=4$
$F^{-}=8$