Silver crystallizes 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.

Given, Edge length of the cell ( a ) = 4.07 × 10 - 8 c m Density d = 10.5 g c m - 3 We also know that: the lattice is FCC type, the number of atoms per unit cell , Z = 4 N A = 6.022 × 10 23 m o le - 1 Using the relation d = Z M a 3 N A Therefore, M = da 3 N A Z = 10 .5 gcm - 3 x 4 .077 × 10 - 8 cm 3 × 6 .022 × 10 23 mole - 1 4 = 107.13 g mole - 1 Therefore, the atomic mass of silver = 107 .13 U

Copper crystallizes into FCC lattice with edge length 3.61 × 10 - 8 c m . Show that the calculated density is in agreement with its measured value of 8.92 g c m - 3 .

Given, Edge length a = 3.61 × 10 - 8 c m . The lattice is of FCC type. Therefore, the number of atoms per unit cell, Z = 4 Atomic mass, M = 63.5 g m o l - 1 We also know that, N A = 6.022 × 10 23 Applying the relation: d = Z M a 3 N A = 4 × 63 .5 g mol - 1 3 .61 × 10 - 8 cm 3 × 6 .022 × 10 23 mol - 1 = 8 . 97 g cm - 3 Hence, the calculated density 8.97 g c m - 3 is in agreement with its measured value which is 8.92 g cm - 3 .

Derive Bragg's equation.

In the following figure of beams, the phases of the beams coincide only when the incident angle equals the reflecting angle. The incident beams are parallel to each other until they reach point Z . When they are at point Z , they strike the surface and travel upwards. At point B the 2nd beam scatters. AB + BC is the distance travelled by the 2nd beam. The extra distance is termed as the integral multiple of the wavelength. nλ = AB + BC We know that AB + BC nλ = 2 AB ------------ (1) Further, D is the hypotenuse of the right ΔABZ . Also AB is the opposite side of θ angle. Hence, AB = d sinθ ------------(2) Substitution of (1) in (2) we get: nλ = 2 d sinθ The equation derived is called Bragg's equation.

Silver crystallizes 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.

Given, Edge length of the cell ( a ) = 4.07 × 10 - 8 c m Density d = 10.5 g c m - 3 We also know that: the lattice is FCC type, the number of atoms per unit cell , Z = 4 N A = 6.022 × 10 23 m o le - 1 Using the relation d = Z M a 3 N A Therefore, M = da 3 N A Z = 10 .5 gcm - 3 x 4 .077 × 10 - 8 cm 3 × 6 .022 × 10 23 mole - 1 4 = 107.13 g mole - 1 Therefore, the atomic mass of silver = 107 .13 U

E M B I B E

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 36

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Earn 100

In terms of band theory, what is the difference between a conductor and a semiconductor?

Important Questions on Solid State

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 37

If

N a C l

is doped with

10^{- 3}

mol

%

of

S r C l_{2}

, what is the concentration of cation vacancies?

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 39

How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 40

Silver crystallizes 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.

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 41

Niobium crystallizes in body-centred cubic structure. If the density is

8.55 g c m^{- 3},

calculate atomic radius of niobium using its atomic mass

93 U

.

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 42

Copper crystallizes into FCC lattice with edge length

3.61 \times 10^{- 8} c m

. Show that the calculated density is in agreement with its measured value of

8.92 g c m^{- 3}

.

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 43

Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 44

Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is $125 p m$ .

What is the length of the side of the unit cell?

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 1 - Solid State>Questions>Q 44

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is $125 pm$ . How many unit cells are there in $1.00 {cm}^{3}$ of aluminium?

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