Kf has nacl structure if the distance between k+ and f- is 269 pm find the density of kf
Answers
We know that D=z×Ma3×NAD=z×Ma3×NA
Given Density = 2.48 g/cm3 z = number of atoms per unit cell.
Given KFis NaCl type So, z will be 4.
M = molar mass of KF = 58.08 g/mol
a = edge length NA = avagadro number = 6.022 x 1023.
Putting all the values in above equation
(1) we get: a3=4×58.82.48×6.023×1023a3=4×58.82.48×6.023×1023=232.321.5×10−24cm3=232.321.5×10−24cm3
a=5.37×10−10cma=5.37×10−10cm = 537 pm
Edge length a = 2(rc+ra)2(rc+ra) rc+ra=a2rc+ra=a2 rc+ra=5372=268pm
Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is 100 Ω. If the resistance of the
same cell when filled with 0.02 mol L–1 KCl solution is 520 Ω, calculate the conductivity and molar
conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 S/m.
21. Calculate the value of Avogadro’s no. of KF (FCC) having density 2.48g/cm3 and distance between K+
and F- is 269pm. KF (At mass K=39, F=19)