1.

By X -ray diffraction methods, the unit length of NaCl is observed to be 0.5627 nm. The density of NaCl is found to be `2.164 cm^(-3)`. What type of defect exists in the crystal ? Calculate the percentage of ` Na^(+) and Cl^(-)` ions missing .

Answer» Calculated density , ` p = (Z xxM)/(a^(3) xx N_(0)) = ( 4 xx 58.5 " g mol"^(-1))/((0.5627 xx 10^(-7) cm)^(3) xx ( 6.022 xx 10^(23) mol^(-1))) = 2.1809 " g cm"^(-3)`
Observed density = ` 2.164 " g cm"^(-3)`
As observed density is less than theoretically calculated valuem this means that some ` Na^(+) and Cl^(-1)` ions are missing from their lattice site , i.e, there is Schottky defect.
Actual formula units of NaCl per unit cell be calculated as follows :
` Z = ( a^(3) xx p xx N_(0))/M = (( 0.5627 xx 10^(-7) cm)^(3) xx ( 2.164 cm^(-3)) xx ( 6. 022 xx 10^(23) mol^(-1)))/( 58.5 " g mol"^(-1)) = 3.968`
Formula unit missing per unit cell = 4 - 3.968 = 0.032
% missing = ` ( 0.032)/4 xx 100 = 0.8 %


Discussion

No Comment Found

Related InterviewSolutions