1.

The number of atoms is 100 g of a fcc crystal with density = 10.0 ` g//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 - d
` p = ( Z xx M)/(a^(3) xx 10^(-30) xx N_(0))`
` or M= ( 10 xx (200)^(3) xx 10^(-30) xx 6 xx 10^(23))/4 = 12`
Thus, 12 g contain = ` N_(0)= 6 xx 10^(23) ` atoms
100 g will contain = `( 6 xx 10^(23))/12 x 100 = 5 xx 10^(24)`


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