The number of atoms in 100 g of a fcc crystal with denstiy = `10.0 g//cm^(3)` and cell edge equal to 200 pm is equal toA. `5xx10^(24)`B. `5xx10^(25)`C. `6xx10^(23)`D. `2xx10^(25)`

The number of atoms in 100 g of a fcc crystal with denstiy = `10.0 g//

1.	The number of atoms in 100 g of a fcc crystal with denstiy = `10.0 g//cm^(3)` and cell edge equal to 200 pm is equal toA. `5xx10^(24)`B. `5xx10^(25)`C. `6xx10^(23)`D. `2xx10^(25)`
Answer» `rho=(Z_(eff)xxMv)/(a^(3)xx10^(-30)xxN_(A))` [For fcc,`Z_(eff)=4`/unit cell] `therefore Mw=(rhoxxa^(3)xx10^(-30)xxN_(A))/(Z_(eff))` `=((10.0gcm^(-3)xx(200"pm")^(3)xx10^(-30)cm^(3)xx6xx10(23)"atoms"))/(4)` `=12g "mol"^(-1)` Thus, 12 g `"mol"^(-1)` contains =`N_(A)` atoms `=6xx10^(23)` atoms `therefore 100g "contains"=(6xx10^(23))/(12)xx100=5xx10^(24)` atoms.

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The number of atoms in 100 g of a fcc crystal with denstiy = `10.0 g//cm^(3)` and cell edge equal to 200 pm is equal toA. `5xx10^(24)`B. `5xx10^(25)`C. `6xx10^(23)`D. `2xx10^(25)`

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