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

Options

A. `3xx10^(25)`
B. `4xx10^(25)`
C. `1xx10^(25)`
D. `2xx10^(25)`

Answer

B

Explanation

`rho = (nM)/(N_(o)a^(3)) " " (a=100p m = 10^(-8)cm)`
`10=(4xxA)/(6.023xx10^(23)xx(10^(-8))^(3))rArr A=(6.023)/(4)=1.508`
`because 1.5 g` contains `6.023xx10^(23)` atoms
`therefore 100 g` contains `= (6.023xx10^(23))/(1.5)xx100=4xx10^(25)`
Hence, (B) is the correct answer.