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An element of atomic mass 40 occurs in f...

An element of atomic mass 40 occurs in fcc structure with a cell edge of 540 pm. Calculate the Avogadro's number if density is `1.7gm//cm^(3)`.

Text Solution

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We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orN_(0)=(ZxxM)/(a^(3)xxrhoxx10^(-30))`
Edge length of the unit cell (a) = 540 pm = 540
No. of atoms per unit cell (Z) = 4
Density of the unit cell `(rho)=1.7gm//cm^(3)`
Atomic mass of the element (M) = `"40 g mol"^(-1)`
`N_(0)=(4xx("40 g mol"^(-1)))/((540)^(3)xx(1.7"g cm"^(-3))xx(10^(-30)cm^(3)))=6.0xx10^(23)"mol"^(-1)`
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