A conductivity cell of cell constant $0.186 c m^{- 1}$ is filled with a $0.0125 M$ solution of an electrolyte and shows a resistance of $100 Ω$ . The molar conductivity of the solution in $S c m^{2} mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in S cm^2 mol^-1, tolerance 0.5.

Answer: 148.8 S cm^2 mol^-1

κ = G^{*} / R = 0.186/100 = 1.86 \times 1 0^{- 3} S c m^{- 1}

Λ_{m} = \frac{1000 κ}{c} = \frac{1000 \times 1.86 \times 1 0 ^{- 3}}{0.0125} = 148.8 S c m^{2} mo l^{- 1}