Question

The metal calcium crystallises in a face centered cubic unit cell with a=0.556 calculate the density of the metal if it contains 0.1% schottky defect

Answer 1

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Answer 2

Answer:The density of the metal $1.5442 g/cm^3$

Explanation:

a = 0.556 nm = $0.576\times 10^{-7} cm$

$\rho=\frac{Z\times M}{N_A\times a^3}$

where,

Z = number of atoms in crystal lattice

M = Atomic mass

a = Edge length of the crystal

$\rho=Density\\N_A=\text{Avogadro's Number}=6.022\times 10^{23}$

For FCC crystal the number of atom in one unit cell = 4

Due to presence of 0.1 % of schottkly defect in the cubic unit cell the value of Z:

$Z=4-4\times \frac{0.1}{100}=3.996$

$\rho=\frac{3.996\times 40 g/mol}{6.022\times 10^{23}\times (0.576\times 10^{-7} cm)^3}=1.5442 g/cm^3$

The density of the metal $1.5442 g/cm^3$