1

AIPMT 2007

If NaCl is doped with 10$-$4 mol% of SrCl2, the concentration of cation vacancies will be (NA = 6.02 $\times$ 1023 mol$-$1)
A
6.02 $\times$ 1016 mol$-$1
B
6.02 $\times$ 1017 mol$-$1
C
6.02 $\times$ 1014 mol$-$1
D
6.02 $\times$ 1015 mol$-$1

Explanation

As each Sr2+ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl2 added.

$\therefore$ Concentration of cation vacancies

= 10–4 mole %

= ${{{{10}^{ - 4}}} \over {100}} \times 6.023 \times {10^{23}}$

= $6.023 \times {10^{23}}$ $\times$ 10-6

= $6.023 \times {10^{17}}$
2

AIPMT 2007

The fraction of total volume occupied by the atoms present in a simple cube is
A
${\pi \over {3\sqrt 2 }}$
B
${\pi \over {4\sqrt 2 }}$
C
${\pi \over 4}$
D
${\pi \over 6}$

Explanation

The maximum properties of the available volume which may be filled by hard sphere in simple cubic arrangement is ${\pi \over 6}$ or 0.52.
3

AIPMT 2006

CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 $\times$ 1023 mol$-$1, the density of CsBr is
A
4.25 g/cm3
B
42.5 g/cm3
C
0.425 g/cm3
D
8.25 g/cm3

Explanation

Density of CsBr = ${{Z \times M} \over {V \times {N_A}}}$

= ${{1 \times 213} \over {{{\left( {436.6 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}$

= 4.25 g/cm3
4

AIPMT 2006

The appearance of colour in solid alkali metal halides is generally due to
A
interstitial positions
B
F.-centres
C
Schottky defect
D
Frenkel defect

Explanation

F-centres are the sites where anions are missing and instead electrons are present. They are responsible for colours.