Notifications
Clear all
0
10/09/2020 6:42 pm
Topic starter
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 10.5 g cm–3, calculate the atomic mass of silver.
1 Answer
0
10/09/2020 6:47 pm
Given, Edge length, a = 4.077 × 10–8 cm
Density, d = 10.5 g cm–3
The given lattice is of fcc type,
Thus the number of atoms per unit cell, z = 4
We also know that NA = 6.022 × 1023/mol
Let M be the atomic mass of silver.
We know, d = zM/a3NA
=> M = da3Na / z = 10.5 x 4.077 × 10–8 x 6.022 × 1023) / 4 = 107.13 g /mol