Q11 of 68 Page 30

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.

Given edge length (a) = 4.07 × 10 -8 cm


 


Density (d) = 10.5gcm-1


 


For fcc number of atoms per unit cell (z) = 4


 


Avagadro number NA = 6.022 × 1023 mol-1


 


We know that, Atomic mass (M)


 


(M)


 


Thus, M


 



 


⇒ M = 106.5 mol-1


 

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