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An element crystallises in a structure h...

An element crystallises in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 400 g of this element contain `48xx10^(23)` atoms.

Text Solution

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The correct Answer is:
`41.67"g cm"^(-3)`
Density of unit cell `(rho)=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))`
According to available data,
Edge length (a) = 200 pm =200
No. of atoms per unit cell (Z) = 4
Atomic mass (M) of the element may be calculated as follows :
`48xx10^(23)` atoms of the element are present in 400 g
`6.022xx10^(23)` atoms of the element are present in `((400g))/(48xx10^(23))xx6.022xx10^(23)=50.18g`
`therefore` Atomic mass (M) of the element = `50.18 "g mol"^(-1)`
Abogadro's number `(N_(0)) = 6.022xx10^(23)"mol"^(-1)`
Density of unit cell `=(4xx("50.18g mol"^(-1)))/((200)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)xx"cm"^(3)))=41.67"g cm"^(-3)`
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