1.

The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal toA. `5 xx 10^(24)`B. `5 xx 10^(25)`C. `6 xx 10^(23)`D. `2 xx 10^(25)`

Answer» Correct Answer - A
`d = (Z xx M)/(N_(A)a^(3))`
`N_(A) = (4 xx 100)/(10 xx (2 xx 10^(-8))^(3)) = 5 xx 10^(24)` (here : 200 Pm `= 2 xx 10^(-8) cm`)


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