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The number of atoms in 100 g of an FCC c...

The number of atoms in 100 g of an FCC crystal with density `d = 10 g//cm^(3)` and cell edge equal to 100 pm, is equal to

A

`4 xx 10^(25)`

B

`3 xx 10^(25)`

C

`2 xx 10^(25)`

D

`1 xx 10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
A
`M = (rho xx a^(3) xx N_(0) xx 10^(-30))/z`
`= (10 xx (100)^(3) xx (6.022 xx 10^(23)) xx 10^(-30))/4 = 15.05`
No. of atoms in 100 g `=(6.022 xx 10^(23))/15.05 xx 100 = 4 xx 10^(25)`
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