Ag crystallizes in Fcc lattice. If edge lenght of the cell is 4. 07×10^8 cm and density is 10.5g cm^3 calculate the atomic mass of Ag
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Density (D), number of atoms per unit cell (Z), atomic mass (M), Avogadro's number (N
0
) and edge length (a) are related as below:
D=
N
0
a
3
ZM
10.5=
6.023×10
23
×(4.07×10
−8
)
3
4M
M=107.9 g/mol
Hence, the atomic mass of silver is 107.9 g/mol
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a×4=bvdxnoiutrfchjhc
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