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Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 xx 10^(-8) cm and its density is 10.5 g cm^(-3). Calculate on this basis the atomic mass of silver. [N_(A)= 6.02 xx 10^(23) mol^(-1)] |
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Answer» Solution :APPLY the RELATION : `=(dxxa^(3)xxN_(A))/(z)` `"Given :d = 10.5 g cm"^(-3), a=4.077xx10^(-8)cm` In fcc LATTICE, z= 4 Substituting the values in the above equation, we get `M=(10.5xx(4.077)^(3)xx10^(-24)xx6.02xx10^(23))/(4)=107.09` Thus, the ATOMIC mass of silver = 107.09u. |
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