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Question: What would be the equivalent conductivity of a cell in which \(0.5\)N salt solution offers a resista...

What would be the equivalent conductivity of a cell in which 0.50.5N salt solution offers a resistance offers 40 ohm whose electrodes are 2 cm apart and 5cm25cm^{2} in area?

A

10ohm−1cmh2eq−110ohm^{- 1}cmh^{2}eq^{- 1}

B

20ohm−1cm2eq−120ohm^{- 1}cm^{2}eq^{- 1}

C

30ohm−1cm2eq−130ohm^{- 1}cm^{2}eq^{- 1}

D

25ohm−1cm2eq25ohm^{- 1}cm^{2}eq

Answer

20ohm−1cm2eq−120ohm^{- 1}cm^{2}eq^{- 1}

Explanation

Solution

k=1R×lA=140×25k = \frac{1}{R} \times \frac{l}{A} = \frac{1}{40} \times \frac{2}{5}

Δeq=k×1000N=140×25×10000.5=20ohm−1cm2eq−1\Delta_{eq} = k \times \frac{1000}{N} = \frac{1}{40} \times \frac{2}{5} \times \frac{1000}{0.5} = 20ohm^{- 1}cm^{2}eq^{- 1}