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. `3 xx 10^(25)`B. `5 xx 10^(24)`C. `1 xx 10^(25)`D. `5.96 xx 10^(-3)`

Answer» Correct Answer - b
`M = (d+ a^(3) xx N_(0))/(n)`
`= (10 xx (200 xx 10^(-3))^(3) xx (6.02 xx 10^(23)))/(4) = 12.04 g`
No of atoms in `100 gm= (6.02 xx 10^(22))/(12.04) xx 100`
`= 5 xx 10^(24)`


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