Silver crystallizes in fcc lattic. If the edge length of the cell is `4.07 xx 10^(-8) cm` and density is `10.5 g cm^(-3)`. Calculate the atomic mass of silver.
Text Solution
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Edge length `(a)=4.077xx10^(-8)cm` `:. ` Volume of unit cell `(a^(3))=(4.077xx10^(-8))^(3)` `=67.77xx10^(-24)cm^(3)` In a `f.c.c.` uinit , there are four atoms per unit cell `(i.e., n=4)` Atomic mass of `Ag` `=("Density" xxAv. no .xx"Volume of unit cell")/(z)` `=(10.5xx6.023xx10^(23)xx67.77xx10^(-24))/(4)` `=107.15`
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