A conductivity cell has cell constant 0.852 cm⁻¹. Filled with 0.001 M AgNO₃ solution, it shows resistance 6530 Ω at 25 °C. The molar conductivity of AgNO₃ at this concentration is approximately:
A131 Ω⁻¹ cm² mol⁻¹
B65.4 Ω⁻¹ cm² mol⁻¹
C13 Ω⁻¹ cm² mol⁻¹
D1305 Ω⁻¹ cm² mol⁻¹
Answer & Solution
Correct answer: A. 131 Ω⁻¹ cm² mol⁻¹
$\kappa = $ cell constant$/R = 0.852/6530 = 1.30 \times 10^{-4}$ Ω⁻¹ cm⁻¹. Molar conductivity $\Lambda = 1000 \kappa/c = 1000(1.30\times10^{-4})/0.001 = 130.5$ Ω⁻¹ cm² mol⁻¹ ≈ 131.