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30/12/2021 6:02 pm
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Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.
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30/12/2021 6:08 pm
It is given that the edge length, a = 4.077 × 10-8 cm
Density, d = 10.5 g cm-3
As the lattice is fcc type, the number of atoms per unit cell, z = 4
We also know that, NA = 6.022 × 1023 mol-1
Using the relation
d = \(\frac{zM}{a^3N_A}\)
M = \(\frac{da^3N_A}{z}\)
= \(\frac{10.5gcm^{-3} \times (4.077 \times 10^{-8})^3 \times 6.022 \times 10^{23}mol^{-1}}{4}\)
= 107.13 gmol-1
Therefore, atomic mass of silver = 107.13 u