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Question

Silver crystallizes in the fcc lattice. If the edge length of the cell is 4.07*10-8 cm and the density is 10.5gcm-3. Calculate the atomic mass of metal?


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Solution

Step 1:Given data:

Silver crystallizing in the fcc lattice

Hence, the number of atoms per unit cell (Z) for Fcc lattice =4

The edge length of the cell is given as (a)=4.07*10-8cm

The density of the unit cell is given as = 10.5gcm-3

Step 2:Formula for density of the unit cell:

The density of a unit cell can be given as:

D=Z×MNA×a3

Where,

Z =The number of atoms per unit cell

a= The edge length of the cell

D=The density of the unit cell

NA=6.023×1023=Avogadro’s number
Step 3: Calculation of the atomic mass of Silver:

By substituting the values in the density formula, we can get:

10.5=4×M(6.023×1023)×(4.07×108)3M=107.9gmol-1

Hence, the atomic mass of the Silver metal is calculated to be 107.9gmol-1


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