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The molar conductivity of a 1.5 M soluti...

The molar conductivity of a 1.5 M solution of an electrolyte is found to be `138.9 S cm^(2) mol^(-1)` . Calculate the conductivity of this solution.

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To calculate the conductivity of the solution, we can use the relationship between molar conductivity (Λm) and conductivity (κ). The formula is given by: \[ \kappa = \Lambda_m \times C \] where: - \( \kappa \) is the conductivity in S cm\(^{-1}\), ...
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Knowledge Check

  • The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm^(2) mol^(-1) . Calculate the conductivity of the solution.

    A
    `1.023 S cm^(-1)`
    B
    `0.026 S cm^(-1)`
    C
    `0.208 S cm^(-1)`
    D
    `0.325 S cm^(-1)`

  • 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 conductivity of a 0.01 M solution of acetic acid at 298 k is 1.65xx10^(-4) S cm ^(-1) . Calculate molar conductivity (wedge_(m)) of the solution.

    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)

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    Calculating molar conductance: The resistance of 0.01 M solution of an electrolyte was found to be 210 ohm at 25^(@)C . Calculate them molar conductance of the colution at 25^(@)C , if the cell constant is 0.88 cm^(-1) Strategy: First conductivity in Sm^(-1) (SI units) and then divide it by molar concentration in mole m^(-3) (SI units) to get molar conductance in Sm^(2) mol^(-1) .

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