1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Calcium crystallises in FCC unit cell with edge length =0.556nm .Calculate its density if it contains 0.1% vacancy defects. (Plz give the solution with explanation)

Open in App
Solution

## Vacancy defect is also sometimes called Schottky defect for non ionic crystals. Here that is the case. The density of the crystal is given by the formula: d = ZM/a^3NA Given data Z for FCC = 4; M (atomic mass) = 40 g/mol; a (inter-atomic distance) = 0.556 nm or 0.556x10^-9 m, Substituting the values in the formula the density would be d = (4*40g)/(0.556x10^-9m)^3*6.023x10^23 = 160/(0.1719x10^-27m3) *6.023x10^23 = 1.546x10^6 g/m3 d = 1.546 g/cm3 Schottky defect reduces the density by 0.1%, assuming that volume remains constant. d’= d( 1- 0.1/100) d’= 0.999d d’= 0.999(1.546g/cm3) d’= 1.544g/cm3 Hope this helps :)

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Zeff
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program