An element with molar mass `2.7xx10^(-2)` kg per mole forms a cubic unit cell with edge length 405 pm. If its density is `2.7xx10^(3)` , what is the nature of the cubic unit cell ?

Explanation

Density = ` p = (Z xx M)/(a^(3) xx N_(A)) or Z = ( p xx a^(3) xx N_(A))/M`
here, M (molar mass of the element )= ` 2.7 xx 10^(-2) " kg mol" ^(-1)`
a( edge lenth) = 405 pm = ` 405 xx 10^(-12) m = 4.05 xx 10^(-10) m`
p (density ) = ` 2.7 xx 10^(3) " kg m"^(-3)`
` N_(A)` (Avogardro's number ) = ` 6.022 xx 10^(23) mol^(-1)`
Substituting these values in expression (i), we get
` Z = ( ( 2.7 xx 10^(3) " kg m"^(-)) ( 4.05xx10^(-10) m)^(3) ( 6.022 xx 10^(23) mol^(-1)))/( 2.7xx 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).