11.139: Estimate the density of potassium chloride using the following ionic radii:

Potassium chloride crystallizes in the rock salt structure.
Estimate the density of potassium chloride using the following ionic radii: r(K^+) = 133pm, r(Cl^-) = 181pm.
d = _____ g/cm^3

**I keep getting the answer to be 1.76 and it's wrong, I don't know what I'm doing wrong. Please help AND please show work, thank you!!

step 1: l = 2r(Cl^-) + 2r(K^+) ===> 2(181pm) + 2(133pm) = 628pm

step 2: V = l^3 ==> (628pm)^3 * ((1cm)^3/(10^10pm)^3) = 2.47673*10^-22 cm^3

step 3: m = (4 formula units/unit cell)*(M/N_A) ==> (4 formula units/unit cell)*(74.55g/1mol)*(1mol/6.022*10^23 formula units) = 4.951976*10^-22 g/unit cell.

step 4: d=m/v ==> d= (4.951976*10^-22g/unit cell)/(2.47673*10^-22cm^3/unit cell) = 1.99940g/cm^3.
Answer: d= 1.99940g/cm^3 OR 2.00g/cm^3


hope this helps!!

Given: KCl, rock salt structure.
Find: density (g/cm^3)
Other: r(K^+)=133pm; r(Cl^-)=181pm
Conceptual Plan: rock salt structure is a face-centered cubic structure with anions at the lattice points and cations in the holes between lattice sites --> assume r = r(Cl^-), BUT M = M(KCl)
..continue..

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...continued...r(K^+), r(Cl^-) --> l and l --> V(pm^3) --> V(cm^3) and, FCC structure --> m then m, V --> d.
Useful formulas:
l = 2r(Cl^-) + 2r(K^+)
V = l^3 ((1cm)^3)/((10^10pm)^3)
m = ((4 formula units)/(unit cell))*(M/N_A)
d = m/V

Hopefully that cleared up some questions. :) good luck!

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