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A metal with a molar mass of 75g*mol^(-1...

A metal with a molar mass of `75g*mol^(-1)` crystallises in a cubic lattice structure with unit cells of an edge length of 5Å. If the density of the metal is `2g*cm^(-3)`, calculate the radius of metal atom.

Text Solution

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We know, `rho=(ZxxM)/(Nxxa^(3))`
Given: `rho=2g*cm^(-3),M=75g*mol^(-1),a=5Å,=5xx10^(-8)cm`
Thus, `2=(75xxZ)/(6.022xx10^(23)xx(5xx10^(-8))^(3))` or, Z = 2
This value of Z (the number of atoms per cubic unit cell), indicates that the metal crystallises in a body-centred cubic lattice. In case of a body-centred cubic unit cell, the relationship between the edge length (a) of the unit cell and the radius ( r) of its atom is, `a=4/sqrt3r`.
`therefore" "r=sqrt3/4a=sqrt3/4xx5xx10^(-8)=2.165xx10^(-8)cm=2.165Å`.
Therefore, the radius of an atom of the metal = 2.165Å.
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