1.

KF has NaCl structure. What is the distance between K^(+) and F^(-) in KF, if its density is 2.48 g cm^(-3) ?

Answer»


Solution :For NACL type structure, Z=4, M = 39 +19=58.0g `mol^(-1)`, d=2.48 g `cm^(-3)`
`a^3=(Z xx M)/(N_A xx d)=(4 xx 58.0)/(6.023 xx 10^(23) xx 2.48) = 15.53 xx 10^(-23) cm^3`:
`a = 5.375 xx 10^(-8)` cm= 537.5 pm
For NaCl type structure, a = `2 xx "interionic distance"`.
`:.` Distance between K and `F^(-)` ions = `(537.5)/(2)` = 268.75 pm


Discussion

No Comment Found

Related InterviewSolutions