The molar conductivity of 0.05 M BaCl₂ solution at 25 °C is 223 Ω⁻¹ cm² mol⁻¹. Its conductivity ($\kappa$) is:
A$0.223$ Ω⁻¹ cm⁻¹
B$4.46 \times 10^{-3}$ Ω⁻¹ cm⁻¹
C$0.01115$ Ω⁻¹ cm⁻¹
D$1.115$ Ω⁻¹ cm⁻¹
Answer & Solution
Correct answer: C. $0.01115$ Ω⁻¹ cm⁻¹
$\kappa = \Lambda \cdot c / 1000 = (223)(0.05)/1000 = 0.01115$ Ω⁻¹ cm⁻¹.
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