In a conductivity cell, the two platinum electrodes, each of area 10 cm2 are fixed 1.5 cm apart. The

Question

In a conductivity cell, the two platinum electrodes, each of area 10 cm2 are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 Ω, the equivalent conductance of the salt solution, in Ω−1 cm2 eq−1, is

Accepted Answer

Conductivity of solution will remain uncharged but the cell constant becomes double.

$\kappa = G \times G^{*} = \frac{1}{50} \times \frac{1.5}{5} = \frac{3}{5} \times 10^{- 2}\text{ oh}\text{m}^{- 1}\text{ c}\text{m}^{- 1} $$$\therefore\Lambda_{eq} = \frac{K}{C} = \frac{\frac{3}{5} \times 10^{- 2}}{0.05 \times 10^{- 3}} = 120\text{ oh}\text{m}^{- 1}\text{ c}\text{m}^{2}\text{ e}\text{q}^{- 1}$$

Question

Options

Solution

More Electrochemistry Questions