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

Given ,
Density of the element, d `= 2.7 xx10^(3) kg m^(-3)`
Molar mass,` M = 2.7 xx 10^(-2) kg mol^(-1)`
Edge length,` a= 405 p m`
`= 405 xx 10^(-12) m`
`= 4.05 xx 10^(-10) m`
Avagadro's number , `N_(A) = 6.022 xx10^(23)mol^(-1)`
`:.d = ( Z xx M )/( a^(3) xx N_(A))`
`rArr Z = ( d xx a^(3) N_(A))/( M )`
`= ( 2.7 xx 10^(3) kg" " m^(-3) ( 4.-5 xx 10^(-10) m ) xx 6.022 xx 10^(23)mol^(-1))/( 2.7 xx 10^(-2) kg' " mol^(-1)`
`= 4.004`
= 4.
This implies that four atoms of the element are present per unit cell. Hence the unit cell is face centred cubic.