Given molar ionic conductivities $\lambda^0$: Ca²⁺ = 104, Cl⁻ = 76.4 (Ω⁻¹ cm² mol⁻¹), the molar conductivity at zero concentration of CaCl₂ is:
A$104 - 76.4 = 27.6$
B$104 + 2(76.4) = 256.8$
C$104 + 76.4 = 180.4$
D$2(104) + 76.4 = 284.4$
Answer & Solution
Correct answer: B. $104 + 2(76.4) = 256.8$
$\Lambda_0$(CaCl₂) = $\lambda^0$(Ca²⁺) + 2$\lambda^0$(Cl⁻) = 104 + 2(76.4) = **256.8** Ω⁻¹ cm² mol⁻¹. The factor of 2 accounts for two chloride ions per formula unit.
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