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The resistance of 0.5 N solution of an e...

The resistance of 0.5 N solution of an electrolyte in a conductivity cell was found to be 25 ohm. Calculate the equivalent conductivity of the solution if the electrodes in the cell are 1.6 cm apart and have an area of `3.2 cm^(2)`

Text Solution

Verified by Experts

The correct Answer is:
`40 ohm^(-1) cm^(2) "equiv"^(-1)`
Cell constant = `(l)/(a) = (1.6)/(3.2) = 0.5 cm^(-1)`
`kappa = (1)/(25) xx 0.5 = 2.0 xx 10^(-2) S cm^(-1)`
`Lambda = (2.0 xx 10^(-2) xx 1000)/(0.5) = 40 S cm^(2) "equiv"^(-1)`
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Knowledge Check

  • The resistance of aN//10KCI solution is 245 Omega . Calculate the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each haveing an area of 7.0 sq.cm .

    A
    `23. 32 cm^2 eq^(-1)`
    B
    `23. 23 S m^2 eq^(-1)`
    C
    `2. 332 S cm^2 eq^(-1)`
    D
    None of these

  • The resistance of an N//10 KCl solution is 245 ohms . Calculate the equivalent conducatnce of the solution id the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq.cm.

    A
    `33.32 S cm^(2)`
    B
    `23.32 S cm^(2)`
    C
    `23.23 S cm^(2)`
    D
    `32.23 S cm^(2)`

  • The resistance of 0.5 N solution of an electroolyte in a conductivity cell was found to be 45 ohms. If the electrodes in the cell are 2.2 cm apart and have an area of 3.8cm^(2) then the equivalent conductance (in Scm^(2)eq^(-1) ) of a solution is

    A
    `25.73`
    B
    `15.75`
    C
    `30.75`
    D
    `35.75`

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