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Question
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.
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Solution
Given: It is given that the edge length, a = 4.07 × 10−8 cm
Density, ρ = 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
Formula: `rho = (Z xx M)/(a^3 xx N_A)`
∴ `M = (rho xx a^3 xx N_A)/Z`
= `(10.5 xx (4.07 xx 10^-8)^3 xx 6.022 xx 10^23)/4`
= `(10.5 xx 67.41 xx 10^-24 xx 6.022 xx 10^23)/4`
= `(4262.4 xx 10^-1)/4`
= `426.24/4`
= 106.56 g mol−1
Therefore, atomic mass of silver is 106.56 g mol−1.
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