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?

Answer

ccp

Explanation

Density, `rho = (Z_(eff) xx Mw)/(a^(3) xx N_(A))` or `Z_(eff) = (rho xx a^(3) N_(A))/(Mw)`
Here, `Mw` (molar mass of the elements)
`= 2.7 xx 10^(-2) kg mol^(-1)`
a (edge length) = `405 pm = 405 xx 10^(-12) m`
`= 4.05 xx 10^(-10)m`
`rho` (density) `= 2.7 xx 10^(3) kg m^(-3)`
`N_(A)` (Avogadro's number) ` = 6.022 xx 10^(23) mol^(-1)`
Substituting these values in expression (i), we get
`Z_(eff) = [((2.7 xx 10^(3) kg m^(-3))(4.05 xx 10^(-10) m^(3)(6.022 xx 10^(23) mol^(-1)))/(2.7 xx 10^(-2) kg mol^(-1))] = 3.99 ~~ 4`
Thus, there are 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centred or cubic close packed (ccp).