Silver crystallises in a fee lattice. If the edge length of the unit cell is 4.07 X 108 cm and density of silver is 10.5 g cm-3, calculate the atomic mass of silver.(NA = 6.02 X 1023 atoms mol-1)

Answered: Silver crystallises in a fee lattice.…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter9: Liquids, Solids, And Materials

Section9.6: Crystalline Solids

Problem 9.9PSP

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Question

Silver crystallises in a fee lattice. If the edge length of the unit cell is 4.07 X 10⁸ cm and density of silver is 10.5 g cm^-3, calculate the atomic mass of silver.(N_A = 6.02 X 10²³ atoms mol^-1)

Expert Solution

Step 1

Note: The edge length of the unit cell should be 4.07 x 10^-8 cm instead of 4.07 x 10⁸ cm in order to get a plausible answer of the molar mass. The solution is provided considering the edge length to be 4.07 x 10^-8 cm.

Step 2

Density (d) of a crystal structure is expressed as:

$d = \frac{Z \cdot M}{N_{a} \cdot a^{3}}$

where, Z is the number of atoms per unit cell, M is the molar mass, N_a is the Avogadro's number and a is the edge length.

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