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The number of atoms in 100 g an fcc crys...

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

A

`1xx10^(25)`

B

`2xx10^(25)`

C

`3xx10^(25)`

D

`4xx10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
D
`d=(ZxxM_(w))/(a^(3)xxN_(A))`
`10=(4xxM_(w))/((10^(-8))xx6xx10^(23))`
`6=4xxM_(w)`
`(6)/(4)=M_(w)rArr1.5g`
`mol = (Mass)(M_(w))`
`(100)/(105)`
"no. of atoms" `=molxxN_(A)`
`(100)/(1.5)xx6xx10^(23)=(6)/(1.5)xx10^(25)=4xx10^(25)`
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