

InterviewSolution
Saved Bookmarks
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 g cm^(-3)`. What type of defect exists in the crystal ?Calculate the percentage of `Na^+` and `Cl^-` ions missing. |
Answer» Calculated density , `rho=(ZxxM)/(a^3xxN_0)=(4xx58.5 "g mol"^(-1))/((0.5627xx10^(-7) cm)^3 xx (6.022xx10^23 "mol"^(-1))=2.1809 " g cm"^(-3)` Observed density =`2.164 "g cm"^(-3)` As observed density is less than theoretically calculated value, this means that some `Na^+` and `Cl^-` ions missing from their lattice site, i.e., there is Schottky defect. Actual formula units of NaCl per unit cell can be calculated as follows : `Z=(a^3xxrhoxxN_0)/M=((0.5627xx10^(-7)cm)^3 xx(2.164 cm^(-3))xx(6.022xx10^23 mol^(-1)))/(58.5 "g mol"^(-1))`=3.968 `therefore` Formula units missing per unit cell=4-3.968=0.032 `therefore` % missing =`0.032/4xx100`=0.8% |
|