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The resistance of a N//10 KCI solution i...

The resistance of a `N//10 KCI` solution is 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq cm.

Text Solution

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Cell constant `= (1)/(a) = (4)/(7) cm^(-1)`.
Specific conductance = conductance `xx` cell constant
`= (1)/("resistance") xx (l)/(a)`
`= (1)/(245) xx (4)/(7) = 2.332 xx 10^(-3) "mho cm"^(-1)`
Equivalent conductance = sp. Cond. `xx` volume (cc) (containing 1 eq. of the substance)
`= 2.332 xx 10^(-3) xx 10000` (for N/10 solution, v = 10,000 cc)
`= 23.32 "mho cm"^(2)`.
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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 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 ci apart and have an area of 3.2 cm^2 ?

    A
    ` 10 S cm^2 eq^(-1)`
    B
    `15 cm^(2) eq^(-1)`
    C
    `20 S cm^(2) eq^(-1)`
    D
    `40 S cm^(2) e`

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    The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equivalent conductivity of the solution if the electrodes in the cell were 2 cm apart and each had an area of 3.5 sq. cm.

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