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The no. of atoms in 100g of a fcc cr...

The no. of atoms in 100g of a fcc crystal with density =10.0 g/cc and edge length as 100 pm is :

A

`3xx10^25`

B

`4xx10^25`

C

`1xx10^25`

D

`2xx10^25`

Text Solution

Verified by Experts

The correct Answer is:
B
Mass of one fcc unit cell =`(4xxM)/N_A`
`rho=(4xxM)/(N_A a^3) " " therefore (4M)/(N_A)=rhoa^3`
Total number of fcc unit cells =`100/(rho a^3)=100/(10xx(10^(-8))^3)=10^25`
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