1.35 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.
1.35 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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1 Answer
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1.35 P= density
A= edge length of the cell
NA= Avogadro Number
Z= no. of atoms in F.C.C unit cell
M= mass of the metal
Edge length of the cell = d = 4.07*10-8 cm
Density = P =10.5g/cm3
No. of unit cell of face centered cubic (F.C.C) lattice is 4, Z=4
Avogadro Number (NA) = 6.022*1023
Mass of silver = M=?
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