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 is2.7xx10^(3) " kg m"^(-3) , what is the nature of the cubic unit cell ?

An element with molar mass 2.7 xx 10^(-2) " kg mol"^(-1) forms a cubic

1.	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 is2.7xx10^(3) " kg m"^(-3) , what is the nature of the cubic unit cell ?
Answer» <P> Solution : 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).