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

Explanation

Density (d) = `(Z xx M)/(a^(3) xx N_(A)) so , Z = ( d xx a^(3) xx N_(A))/(M)`
Given , M = `2.7 xx 10^(-2) kg mol^(-1)`
`a = 405 ` pm = `405 xx 10^(-12) m = 4.05 xx 10^(-10)` m
`d = 2.7 xx 10^(3) kgm^(-3)`
`N_(A) = 6.022 xx 10^(23) mol^(-1)` .
`Z = ((2.7 xx 10^(3) kg m^(-3)) (4.05 xx 10^(-10) m)^(3) xx (6.022 xx 10^(23) mol^(-1)))/((2.7 xx 10^(-2) kg mol^(-1)))`
`=3.99 = 4 `
Since , there are four atoms per unit cell , the cubic unit cell must be face - centred.