An element X (atomic weight = 24 gm mole ) forms a face centred cubic lattice. If the edge length of the lattice is 4 × 10 - 8 cm and the observed density is 2 . 40 × 10 3 kg m 3 , calculate the percentage occupancy of lattice points by X element. (Given N A = 6 × 10 23

The density of a unit cell = Z × M N A × a 3 Whereas: Z = the number of atoms in a unit cell. M = molar mass N A = Avogadro's number a = the edge length of the unit cell Putting all the given values: Ideal Density = 4 × 24 × 10 - 3 6 × 10 23 × 4 × 10 - 10 3 = 2 . 5 × 10 3 kg m - 3 % occupancy = Observed density Ideal density × 100 = 2 . 4 × 10 3 2 . 5 × 10 3 × 100 = 96 %

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10 - 8 c m and density is 10.5 g c m - 3 , calculate the atomic mass of silver. N A = 6 . 022 X 10 23 mol - 1

In FCC lattice, Z = 4 Edge length (a) = 4 . 07 × 10 - 8 cm Density (d) = 10 . 5 g / cm 3 Avogadro number ( N A ) = 6 . 022 × 10 23 mol - 1 d = z . M a 3 . N A Atomic mass (M) = d . ( a ) 3 . ( N A ) z = 10 . 5 × 4 . 07 × 10 - 8 3 × ( 6 . 022 × 10 23 ) 4 = 106 . 57 Therefore, the atomic mass of silver is 106 . 57 g mol - 1 .

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The co-ordination number of FCC structure for metals is $12$ , since

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Important Questions on Solid State

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-2>Q 14

A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge. The radius of Mn ion is

0.65 \overset{°}{A}

and that of

F^{-}

ion is

1.36 \overset{°}{A}

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-3>Q 15

Cesium chloride forms a body centered cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is

412 pm

and

{Cl}^{-}

ion has a radius of

181 pm .

Calculate the radius of

{Cs}^{+}

ion.

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-3>Q 22

A

compound formed by elements

X & Y,

Crystallizes in a cubic structure, where

X

is at the corners of the cube and

Y

is at six face centers. What is the formula of the compound? If side length is

5 \overset{o}{A}

, estimate the density of the solid assuming atomic weight of

X

and

Y

60

and

90

respectively.

N_{A} = 6.022 X 10^{23}

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-3>Q 23

An element

X

(atomic weight

= \frac{24 gm}{mole}

) forms a face centred cubic lattice. If the edge length of the lattice is

4 \times 10^{- 8} cm

and the observed density is

\frac{2.40 \times 10^{3} kg}{m^{3}},

calculate the percentage occupancy of lattice points by

X

element. (Given

N_{A} = 6 \times 10^{23})

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-4>Q 1

The element chromium exists as a

bcc

lattice whose unit cell edge is

2.88 \overset{\circ}{A} .

The density of chromium is

\frac{7.20 g}{cc} .

How many atom does

52.0 g

of chromium contain? Atomic mass of Cr =

52

N_{A} = 6.022 X 10^{23} {mol}^{- 1}

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-4>Q 2

The edge length of the unit cell of

KCl (NaCl

like structure ;

fcc)

6.28 \overset{\circ}{A} .

Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion.

(r_{Cl} = 1.8173 \overset{\circ}{A}) .

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-4>Q 3

Silver crystallises in fcc lattice. If edge length of the cell is

4.07 \times 10^{- 8} c m

and density is

10.5 g {c m}^{- 3}

, calculate the atomic mass of silver.

N_{A} = 6.022 X 10^{23} {mol}^{- 1}

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Beta Question Bank for Engineering: Chemistry>Chapter 7 - Solid State>EXERCISE-4>Q 4

If coordination number of primitive cubic arrangement is

' x', bcc

arrangement is

' y'

and

fcc

arrangement is

' z'

then give

\frac{x \times z}{y} .

The co-ordination number of FCC structure for metals is 12, since

Important Questions on Solid State

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The co-ordination number of FCC structure for metals is $12$ , since