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In a conductivity cell, the two platinum...

In a conductivity cell, the two platinum electrodes, each of area 10sq. Cm 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 ohms, find equivalent conductance of the salt solution in `Omega^(-1)cm^(2)eq^(-1)`.

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The correct Answer is:
`120mhocm^(2)eq^(-1)`
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RESONANCE-ELECTROCHEMISRY-Exercise
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  2. The resistance of a solution A is 50 ohm and that of solution B is 100...

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  3. In a conductivity cell, the two platinum electrodes, each of area 10sq...

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  4. The equivalent conductance of 0.10 N solution of MgCI(2) is 97.1 mho c...

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  9. Calculate K(a) of acetic acid it its 0.05N solution has equivalent con...

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  10. The sp. Cond. Of a saturated solution of AgCl at 25^(@)C after subtrac...

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  11. In a galvanic cell

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  12. Which of the following statements is wrong about galvanic cells?

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  13. Which of the following is/are function(s) of salt-bridge?

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  14. Salt bridge contains

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  15. E^(0) for F(2)+2eto2F^(-) is 2.8V,E^(0) for (1)/(2)F(2)toF^(-) is

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  16. Consider the cell potentials E(Mg^(2+)|Mg)^(0)=-2.37V and E(Fe^(3+)|Fe...

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  17. If a spoon of copper metal is placed in a solution of ferrous sulphate...

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  18. The position of some metals in the electrochemical series in dectreasi...

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  19. Given : E^(c-).(Ag^(o+)|Ag)=0.80V, E^(c-).(Mg^(2+)|Mg)=-2.37V, E^(...

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  20. For Zn^(2+) //Zn, E^@ =- 0. 76 V, for Ag^+ //Af E^@ =0.799 V. The cor...

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