If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.023 x 1023 mol-1)
(a) 6.023 x 1015 mol-1 (b) 6.023 x 1016 mol-1
(c) 6.023 x 1017 mol-1 (d) 6.023 x 1014 mol-1
(c) Doping of NaCl with 10-4 mol % of SrCl2 means, 100 moles of NaCl are doped with 10-4 mol of SrCl2.
... 1 mol of NaCl is doped with
SrCl2 = 10-4/100 = 10-6 mole
As each Sr2+ ion introduces one cation vacancy.
... Concentration of cation vacancies
= 10-6 mol/mol of NaCl
= 10-6 x 6.023 x 1023 mol-1
= 6.023 x 1017 mol-1
CsCl crystallises in body-centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct?
(a) rCs+ + rCl- = 3a
(b) rCs+ + rCl- = 3a/2
(c) rCs+ + rCl- = (/2)*a
(d) rCs+ + rCl- =
(C) For bcc if edge length is a, then
rCs+ + rCl- = (/2)*a
A solid PQ have rock salt type structure in which Q atoms are at the corners of the unit cell. If the body centred atoms in all the units cells are missing, the resulting stoichiometry will be
(C) In rock salt structure, atoms at the corner are also present at the face centres and hence
ZQ=8 x 1/8 + 6 x 1/2 =4
And, the atoms at body centre also present at edge centres. But from question, body centres are missing and hence
ZP = 12 x 1/4 = 3
Formula of resulting solid - P3Q4.
Lithium borohydride (LiBH4), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are:- a = 6.81 Å , b = 4.43 Å ,c = 717 Å. If the molar mass of LiBH4 is
21.76 g mol-1.The density of the crystal is:-
(A) 0.668 gcm-3
(B) 0.585 gcm2
(C) 1.23 gcm-3
(D) None of these
The arrangement of X- ions around A+ ion in solid AX is given in the figure (not drawn to scale).
If the radius of X- is 250 pm,the radius of A+ is:
(d) 57 pm
(a) The figure represents an octahedral void
In an f.c.c. unit cell. atoms are numbered as shown below.The atoms not touching each other are (Atom numbered 3 is face centre of front face).
(C) Atoms along one edge or at corners do not touch each other in fee cell.