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An element with molar mass 2.7 xx 10^-2 ...

An element with molar mass `2.7 xx 10^-2 kg mol^(-1)` forms a cubic unit cell with edge length 405pm.If its density is `2.7 xx 10^(3) kg^(-3)`, what is the nature of the cubic unit cell?

Text Solution

Verified by Experts

Density `rho = (ZxxM)/(a^(3)xxN_(A)) or Z= (rhoxxa^(3)xxN_(A))/(M)" " cdots(i)`
Here, M (molar mass of the element) `= 2.7xx10^(-2) kg per mole`
a (edge length ) = 405 pm = `404 xx 10^(-12)= 4.05 xx 10^(-10)m, rho ("density") = 2.7 xx 10^(3) "kgm"^(-3)`
`N_(A)` ( Avogadro's num ber ) = `6.022 xx 10^(23) "mol"^(-1)`
Substituting these values in expression (i), we get :
`Z = ((2.7xx10^(3)) "kgm"^(-3).(4.05xx10^(-10)m)^(3)(6.022xx10^(23)mol^(-1)))/(2.7xx10^(-2)"kgmol"^(-1))= 3.994`
As number of particles is 4, hence lattice is fcc.
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Knowledge Check

  • An element with density 2.8 g cm^(-3) forms a fee unit cell with edge length 4 xx 10^(-8) cm. Calculate the molar mass of the element.

    A
    `42g "mol"^(-1)`
    B
    `27g "mol"^(-1)`
    C
    `36g "mol"^(-1)`
    D
    `28g "mol"^(-1)`