Example calculation for a sample of CuSO₄.5H₂O

R₀ = -28
R = 592
l  = 2.1 cm
m = 0.238 g
C = 1.089
T = 293 K

χ_g  =  C(R-R₀)l  = 1.089 x 620 x 2.1  =  5.957 x 10^-6
            10⁹m             0.238 x 10⁹


Molecular weight of CuSO₄.5H₂O = 249.7

Hence  χ_m = 5.957 x 10^-8 x 249.7 = 1.488 x 10^-3 cm³mol^-1

Diamagnetic corrections:  -12.8 x 10^-6 (Cu²⁺) + -40 x 10^-6 (SO₄^2-) + 5 x -13 x 10^-6 (H₂O) = -117.8 x 10^-6

Therefore χ_m' =  1.488 x 10^-3 - (-117.8 x 10^-6) = 1.606 x 10^-3 cm³mol^-1

Effective magnetic moment = (8 χ'm T)^½ = (8 x 1.606 x 10^-3 x 293)^½ = 1.94 BM


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