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A metal of density 7.5xx10^(3) kg m^(-3)...

A metal of density `7.5xx10^(3) kg m^(-3)` has an fcc crystal structure with lattice parameter a = 400 pm. Calculate the number of unit cells present in 0.015 kg of the metal.

A

`6.250xx10^(22)`

B

`3.125xx10^(23)`

C

`3.125xx10^(22)`

D

`1.563xx10^(22)`

Text Solution

Verified by Experts

The correct Answer is:
C
The volume available = `0.015/(7.5xx10^(3))`, (Number of unit cells) `xx(400xx10^(-12))^(3)=0.015/(7.5xx10^(3))`
`rArr` Number of unit cell = `3.125xx10^(22)`
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