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The number of atoms in 100 g of an fcc c...

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 to

A

`5 xx 10^(24`

B

`5 xx 10^(25)`

C

`6 xx 10^(23)`

D

`2 xx 10^(25)`

Text Solution

Verified by Experts

The correct Answer is:
A
`rho=(Z_(eff)xxMw)/(a^(3)xx10^(-30)xxN_(A))`, [For fcc, `Z_(eff) = 4//"unit cell"`]
`:. Mw = (rho xx a^(3) xx 10^(-30) xx N_(A))/(Z_(eff))`
`((10.0g cm^(-3) xx (200 "pm")^(3) xx 10^(-30)cm^(3)xx6xx10^(23)"atoms"))/(4)`
`= 12 g mol^(-1)`
Thus, `12 g mol^(-1)` contains `= N_(A)` atoms `= 6 xx 10^(23)` atoms
`:. 100 g` contains `= (6 xx 10^(23))/(12) xx 100 = 5 xx 10^(24)` atoms
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