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1

NEET 2013

MCQ (Single Correct Answer)
A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm$$-$$3. The molar mass of the metal is
(NA Avogadro's constant = 6.02 $$ \times $$ 1023 mol$$-$$1)
A
27 g mol$$-$$1
B
20 g mol$$-$$1
C
40 g mol$$-$$1
D
30 g mol$$-$$1

Explanation

for fcc Z = 4.

d = $${{ZM} \over {{N_A}{a^3}}}$$

$$ \Rightarrow $$ M = $${{d{N_A}{a^3}} \over Z}$$

= $${{2.72 \times 6.023 \times {{10}^{23}} \times {{\left( {404 \times {{10}^{ - 10}}} \right)}^3}} \over 4}$$

= 27 g mol–1
2

AIPMT 2012 Mains

MCQ (Single Correct Answer)
Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are cooupied by a monovalent metal B. The formula of the oxide is
A
ABO2
B
A2BO2
C
A2B3O4
D
AB2O2

Explanation

Number of atoms in cubic close packing = 4 = O2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A2+ ions = 8 $$ \times $$ $${1 \over 4}$$ = 2

Number of octahedral voids = Number of B+ ions = N = 4

Ratio of ions will be,

O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB2O2
3

AIPMT 2012 Prelims

MCQ (Single Correct Answer)
The number of octahedral void(s) per atom present in a cubic close-packed structure is
A
1
B
3
C
2
D
4

Explanation

Number of octahedral voids is same as number of atoms.
4

AIPMT 2012 Prelims

MCQ (Single Correct Answer)
A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
A
288 pm
B
408 pm
C
144 pm
D
204 pm

Explanation

Given $$a$$ = 408 pm

For the face centred cubic structure

r = $${a \over {2\sqrt 2 }}$$ = $${{408} \over {2\sqrt 2 }}$$ = 144 pm

$$ \therefore $$ Diameter = 2r = 2$$ \times $$144 = 288 pm

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