Silver crystallises in fcc structure with edge length of unit cell, 4 . 07 × 10 - 8 cm and if density of metallic silver is 10 . 5 g cm - 3 . Calculate the molecular mass of silver.

Volume of unit cell = a 3 = ( 4 . 07 × 10 - 8 cm ) 3 = 6 . 74 × 10 - 23 cm 3 Density = Mass of unit cell Volume of unit cell Therefore, 10 . 5 g cm - 3 = Mass of unit cell 6 . 74 × 10 - 23 cm 3 Therefore, mass of unit cell = 10 . 5 × 6 . 74 × 10 - 23 = 70 . 8 × 10 - 23 g Since, the unit cell is fcc type, the number of atoms per unit cell is four. Therefore, mass of one atom = 70 . 8 × 10 - 23 4 = 17 . 7 × 10 - 23 g atom - 1 Atomic mass = Mass of one atom × Avogadro number = 17 . 7 × 10 - 23 g atom - 1 × 6 . 023 × 10 23 atom mol - 1 = 106 . 6 g mol - 1 Since, silver is monoatomic, molecular mass of silver is equal to atomic mass of silver. Therefore, molecular mass of silver = 106 . 6 g mol - 1

E M B I B E

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 17

HARD

12th Tamil Nadu Board

IMPORTANT

Earn 100

The length of a unit cell in the nickel crystal is $0.352 nm$ . Diffraction of $X$ -rays of $0.154 nm$ wavelength from a nickel crystal occurs at $22.2 °, 25.9 °$ , and $38.2 °$ . Show that these data are consistent with a face-centered cubic crystal structure.

Important Questions on Solid State

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 18

Explain with the help of a diagram.

Seven types of unit cells

HARD

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 19

Silver crystallises in fcc structure with edge length of unit cell,

4.07 \times 10^{- 8} cm

and if density of metallic silver is

d_{(2, 0, 0)} : d_{(1, 1, 0)} : d_{(1, 1, 1)} = \frac{a}{2} : \frac{a}{2 \sqrt{2}} : \frac{a}{\sqrt{3}} = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}

. Calculate the molecular mass of silver.

HARD

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 20

Unit cell of iron crystal has edge length of

288 pm

and density of

7.86 g {cm}^{- 3}

. Determine the type of crystal lattice

(Fe = 56)

.

HARD

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 21

An atom crystallises in fcc crystal lattice and has a density of

10 g {cm}^{- 3}

with unit cell edge length of

100 pm

. Calculate number of atoms present in

1 g

of crystal.

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 22

An element A and B constitute bcc type crystalline structure. Element A occupies body centre position and B is at the corners of cube. What is the formula of the compound? What are the coordination numbers of A and B?

EASY

12th Tamil Nadu Board

IMPORTANT

Chemistry Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Solid State>Exercise>Q 23

Atoms