A conductivity cell has electrodes of area $1.5 c m^{2}$ placed $0.6 cm$ apart. A 0.05 M NaCl solution in this cell has a resistance of $80 Ω$ . The molar conductivity of the solution in $S c m^{2} mo l^{- 1}$ is.

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

Answer: 100 S cm^2 mol^-1

Cell constant

= l / A = 0.6/1.5 = 0.4 c m^{- 1}

κ = 0.4/80 = 5 \times 1 0^{- 3} S c m^{- 1}

Λ_{m} = \frac{1000 κ}{c} = \frac{1000 \times 5 \times 1 0 ^{- 3}}{0.05} = 100 S c m^{2} mo l^{- 1}