A 0.05 N solution of a salt occupying a volume between two platinum electrodes separated by a distance of 1.72 cm and having an area of `4.5 cm^(2)` h

Question

A 0.05 N solution of a salt occupying a volume between two platinum electrodes separated by a distance of 1.72 cm and having an area of `4.5 cm^(2)` h

A 0.05 N solution of a salt occupying a volume between two platinum electrodes separated by a distance of 1.72 cm and having an area of `4.5 cm^(2)` has a resistance of 250 ohm. Calculate the equivalent conductance of the solution.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Specific conductance = Conductance `xx` Cell constant
`kappa=Cxxl/A`
`=1/Rxxl/A`
`=1/250xx1.72/4.5`
`=1.5288xx10^(-3) ohm^(-1) cm^(-1)`
`Lambda_(e)=kappaxx1000/N`
`=1.5288xx10^(-3) xx1000/0.05`
`=30.56 ohm^(-1) cm^(2) eq^(-1)`

4 views Answered 2023-02-05 15:29