Science > Chemistry > solid State > Numerical problems on density of Solid

In this article, us shall examine to solve numerical troubles to uncover the thickness of solid, sheet length and volume of the unit cell.

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Example – 01:

Silver crystallizes in challenge centred cubic structure. Theedge size of a unit cell is uncovered to it is in 408.7 pm. Calculate thedensityof silver. (Ag = 108 gmol-1)

Given: The edge size of the unit cell = a = 408.7 afternoon = 408.7 x10-10cm, atomic mass of silver- = M = 108 g mol-1,Avogadro’s number N = 6.022 x 1023 mol-1. Type of crystalstructure = confront centred cubic

ToFind: thickness of silver- =?

Solution:

The variety of atoms in the unit cabinet of a challenge centred cubicstructure is n = 4 Volume the unit cabinet = a3 = (4.30x 10-8cm) 3 = (4.3) 3 x 10-24 cm3 = 79.4x 10-24 cm3 The density of salt is 0.962 g cm-3.

Ans: Density of sodium = 0.962 g/cm3

Example – 06:

copper crystallizes in fcc form unit cell. The edge size of a unit cell is 360.8 pm. The density of metallic copper is 8.92 g cm-3. Determine the atomic mass that copper.

Given: The edge length of the unit cabinet = a = 360.8 pm = 360.8 x10-10 cm = 3.608 x 10-8 cm, density of copper= 8.92 g cm-3, Avogadro’s number N = 6.022 x 1023 mol-1.Type of crystal framework = fcc

ToFind: the atomic mass of copper =?

Solution:

The variety of atoms in the unit cabinet of a face centred cubicstructure is n = 4

Ans: The atom mass that copper is 63.07 g mol-1.

Example – 07:

Silver crystallises in fcc kind unit cell. The leaf lengthof aunit cell is 4.07 10-8 cm. The thickness of metallic silveris 10.5 g cm-3. Recognize atomic fixed of silver.

Given: The edge size of the unit cell = a = 4.07 x 10-8cm, density of silverr = 10.5 g cm-3, Avogadro’s number N = 6.022 x1023 mol-1. Form of crystal structure = fcc

ToFind: atom mass of silver =?

Solution:

The number of atoms in the unit cabinet of a challenge centred cubicstructure is n = 4

Ans: The atomic mass of silver- is 106.6 g mol-1

Example – 08:

Face centred cubic crystal lattice the copper has actually a densityof 8.966 g cm-3. Calculate the volume of the unit cell. Giventhemolar mass of copper 63.5 g/mol.

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Given: Molar fixed of copper = 63.5 g mol-1, density of copper = 8.966 g cm-3, Avogadro’s number N = 6.022 x 1023 mol-1. Type of crystal framework = fcc

To Find: Volume the unit cell =?

Solution:

The number of atoms in the unit cabinet of a confront centred cubicstructure is n = 4

Ans: The volume that the unit cell is 4.704 x 10-23 cm3

Example – 09:

Silver crystallizes together fcc structure. If the density of silver- is 10.51 g cm-3. Calculation the volume of a unit cell. Given: molar massive of silver 108 g/mol.

Given: Molar mass of silver = 108 g mol-1, density ofsilver = 10.51 g cm-3, Avogadro’s number N = 6.022 x 1023mol-1. Type of crystal structure = fcc

ToFind: Volume the unit cabinet =?

Solution:

The number of atoms in the unit cell of a confront centred cubicstructure is n = 4

Ans: The volume of the unit cell is 6,825 x 10-23 cm3

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## 2 replies on “Numerical troubles on thickness of Solid” Choudhary Himanshusays:
February 5, 2021 at 9:59 am

Good questions. Really assisted a lot