KF crystallizes in fcc structure like sodium chloride. Calculate the distance between `K^(+)` and `F^(-)` in KF. ( Given `:` density of KF is `2.48 g cm^(-3)` )

Explanation

We know that D `=(Z xxM )/ ( a^(3) xx N_(A))`
Given Density `= 2.48 g cm^(-3)`
Z = No. of atoms per unit cell
Given KF is NaCl typeso Z will be 4
M = Molar masses of KF
`= 58.08 g//mol`
a =edge length NA
= Avogadro number
Putting all the values in above questions.
(1) We get `d^(3)= ( 4 xx 58.8 ) /( 2.48 xx 6.023 xx 10^(23))`
`= (232.32)/( 1.5 xx 10^(-24) ) cm^(3)`
`a = 5.37 xx 10^(-10) cm`
a = 537 pm
Edge length `a= 2 (rc +ra ) rc + ra= ( 9)/(2)`
`= rc + ra =( 537)/( 2)`
= 268 pm