Limiting molar ionic conductivities: lambda deg(Na+) = 50.1 and lambda deg(Cl-) = 76.3 S cm^2 mol^-1. The transport number of Cl- in a very dilute NaCl solution (three decimals) is

JEE Main levelNumerical answerDifficulty 2 of 3ec217

Limiting molar ionic conductivities: $λ^{\circ} (Na X^{+}) = 50.1$ and $λ^{\circ} (Cl X^{-}) = 76.3 S c m^{2} mo l^{- 1}$ . The transport number of $Cl X^{-}$ in a very dilute NaCl solution (three decimals) is

Numerical answer type, tolerance 0.003.

At infinite dilution the transport number equals the fraction of the total ionic conductivity:

t_{-} = \frac{λ _{-}^{\circ}}{λ _{+}^{\circ} + λ _{-}^{\circ}}

t (Cl X^{-}) = \frac{76.3}{50.1 + 76.3} = \frac{76.3}{126.4} = 0.604

. Correspondingly

t (Na X^{+}) = 0.396