Question

Calcium crystallises in a face centred cubic unit cell .The volume of unit cell is $1.72\times {10}^{-22}c{m}^3$. What will be its density $\left(ingc{m}^{-3}\right)$ if it contains $0.5\%$ of Schottky defect?

• A. $0.44$
• B. $3.57$
• C. $1.53$
• D. $2.88$

Answer 1

The correct option is C $1.53$

The number of atom per fcc unit cell is 4
Schottky defect = 0.5%

Schottky defect arises when a pair of one cation and one anion of equal valency is missing from an ionic crystal, thus a pair of vacant sites are formed. These ion–pair vacancies decrease density of the solid.

Number of calcium atom per unit cell with schottky defect is,
$Z=4(1-\frac{0.5}{100})$
$Z=3.98$
$\text{Density,}\rho =\frac{Z\times M}{N_A\times V}$
where,
$Z$ is number of atoms in unit cell
$N_A$ is avagadro number
$M$ is molar mass of element
$V$ is volume of unit cell

$\text{Density,}\rho =\frac{3.98\times 40}{6.023\times {10}^{23}\times 1.72\times {10}^{-22}}$

$\text{Density,}\rho =1.53gc{m}^{-3}$