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An element with density 11.2"g cm"^(-3) ...

An element with density `11.2"g cm"^(-3)` forms a fcc lattice with edge length of `4xx10^(-8)` cm. Calculate the atomic mass of the element.

Text Solution

Verified by Experts

`M=(rhoxxa^(3)xxN_(0)xx10^(-30))/(Z)`
According to available data
Edge length (a) = `4xx10^(-8)cm=400xx10^(-10)cm=400"pm"=400`
No. of atoms per unit cell = 4
Density of the element `=11.2"g cm"^(-3)`
Avogadro's No`(N_(0))=6.022xx10^(23)"mol"^(-1)`
Atomic mas of the element (M) `=((11.2"g cm"^(-3))xx(400)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)"cm"^(3)))/(4)`
`=107.9"g mol"^(-1)=107.9 u`.
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