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NEET-XII-Chemistry

03: Electrochemistry

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  • #11
    Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
    Ans : Given, κ = 7.896 × 10-5 S m-1

    c = 0.00241 mol L-1

    Then, molar conductivity,



    = 32.76S cm2 mol-1

    Again, = 390.5 S cm2 mol-1

    Now,

    = 0.084

    Dissociation constant,



    = 1.86 × 10-5 mol L-1