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Q. The number of atoms present in 100 g ...

Q. The number of atoms present in 100 g of a bcc crystal (density=10 g `cm^(-3)`) having cell edge 200 pm is

A

`1xx10^(25)`

B

`1xx10^(24)`

C

`2xx10^(24)`

D

`2xx10^(26)`

Text Solution

Verified by Experts

The correct Answer is:
C
The density is given by `rho=(zxxM)/(a^(3)xxN_(A))`
Therefore, `12.5=(2xxM)/(200xx200xx200xx10^(-30)xx6.02xx6.02xx10^(23))implies M=30.1g`
Now, 30.1 g contains `6.02xx10^(23)` atoms. So, 100g will contain
`(6.02xx10^(23)xx100)/(30.1)=2xx10^(24)` atoms
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Density of a unit cell is the same as the density of the substance. So, if the density of the substance is known, we can calculate the number of atoms or dimensions of the unit cell. The density of the unit cell is related to its formula mass (M), number of atoms per unit cell (z), edge length (a in cm), and Avogadro's constant N_(A) , as rho=(zxxM)/(a^(3)xxN_(A)) Q. An element X crystallizes in a structure having an fcc unit cell of an edge 100 pm. if 24 g of the element contains 24xx10^(23) atoms, the density is