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The number of atoms is 100 g of a fcc cr...

The number of atoms is 100 g of a fcc crystal with density = 10.0 ` g//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:
d
` p = ( Z xx M)/(a^(3) xx 10^(-30) xx N_(0))`
` or M= ( 10 xx (200)^(3) xx 10^(-30) xx 6 xx 10^(23))/4 = 12`
Thus, 12 g contain = ` N_(0)= 6 xx 10^(23) ` atoms
100 g will contain = `( 6 xx 10^(23))/12 x 100 = 5 xx 10^(24)`
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