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. `4 xx 10^(25)`B. `3 xx 10^(25)`C. `2 xx 10^(25)`D. `1 xx 10^(25)`

Answer» Correct Answer - a
`M= ( rho xx a^(3) xx N_(0) xx 10^(-30))/(z)`
`= (10 xx (100)^(2) (6.02 xx 10^(23))xx 10^(-30))/(4) = 15.05`
No of atoms in `100 g = (6.02 xx 10^(23))/(15.05) xx 100 = 4 xx 10^(23)`


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