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A metal of density 7.5 xx 10^(3) kg m^(-...

A metal of density `7.5 xx 10^(3) kg m^(-3)` has an fcc crystal structure with lattice parameter `a = 400 pm`. Calculater the number of unit cells present in `0.015 kg` of metal.

A

`6.250xx10^(22)`

B

`3.125xx10^(22) `

C

`3.125xx10^(22)`

D

`1.563xx10^(22)`

Text Solution

Verified by Experts

The volume available `=(0.015)/(7.5xx10^(3))`
(no. of unit cells) `xx [400xx10^(-12)]^(3)= (0.015)/(7.5xx10^(3)) implies` no. of unit cells `= 3.125 xx10^(22)`
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