1.

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 toA. `3xx10^(25)`B. `5xx10^(24)`C. `1xx10^(25)`D. `2xx10^(25)`

Answer» Correct Answer - 2
`a=200 p m =200xx10^(-10)cm=2xx10^(-8)cm`
volume `=(2xx10^(-8))^(3)cm^(3)`
No. of atoms `=(ZxxA)/(d xx a^(3))=(4xx100)/(10xx(2xx10^(-8))^(3))=5xx10^(24)`


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