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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 `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal to

A

`5 xx 10^(24)`

B

`5 xx 10^(25)`

C

`6 xx 10^(23)`

D

`2 xx 10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
A
`d = (Z xx M)/(N_(A)a^(3))`
`N_(A) = (4 xx 100)/(10 xx (2 xx 10^(-8))^(3)) = 5 xx 10^(24)` (here : 200 Pm `= 2 xx 10^(-8) cm`)
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