15.) 508X6.02 /mol
1`In the face centered cubic arrangement of A and B atoms where A atoms are at the corner of the unit cell and B atoms at the face centres. One of the A atom is missing from one corner in the unit cell. What is the simplest formula of the compound?
2`In Chromium(III) Chloride, CrCl3, chloride ions have cubicclosepacked arrangement and Cr(III)ions are present in the octahedral holes. What is the fraction of octahedral holes occupied? What is the fraction of total number of holes occupied?
3`Iron changes its crystal structure from body centred to cubic close packed structure when heated to 916°C. Calculate the ratio of the density of the BCC crystal to that of CCP crystal. Assume that the metallic radius of the atom does not change.
4`A compound forms hexagonal close packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
5`The concentration of cation vacancies in NaCl crystal doped with CdCl2 is found to be 6.02x1016 mol-1. What is the concentration of CdCl2 added to it?
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5 Answers
334.
No of oct voids= 0.5*NA
no of tetra voids =2*0.5*NA
total no of voids= 1.5*NA
1061.) A7B24 i think
A occupies corners. so each atom of A has contribution of 1/8 (8 unit cells are attached to one corner atom) to the unit cell and there are 7 A atoms in unit cell [1 is missing] so total contribution of A atoms to one unit cell is 7*1/8 = 7/8
B occupies face centres hence its contribution to single unit cell is 1/2 (as each face centre cell is shared by two faces). As there are 6 B atoms in one unit cell. so total contribution of B atoms to one unit cell is 6*1/2 = 3
So, formula of unit cell is A7/8B3 <=> A7B24
