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The density of KBr is "2.75 g/cm"^3 and ...

The density of KBr is `"2.75 g/cm"^3` and edge length of unit cell is 654 pm (K=39 , Br=80) then what are the features of this unit cell ?

Text Solution

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For KBr, `M = 39+80=119`
`a=(654xx10^(-10))cm`
`d=2.75g//cm^(3)`
`therefore rho =(Z xx M)/(N_(0)xx a^(3))` or `Z = (rho xx N_(0)xx a^(3))/(M)`
`Z = (2.75xx 6.02xx10^(23)xx(654xx10^(-10))^(3))/(119)`
On solving, Z = 3.89 = 4
Hence, it is a fcc lattice.
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