12th Class Chemistry Electrochemistry / विद्युत् रसायन

  • question_answer 26)
    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 390.5S cm2 mol?1, what is its dissociation constant?

    Answer:

    Specific conductance or donductivity = 7.896 × 10?5 S cm?1 C = 0.00241 M = 32.76 S cm2 mol?1. Dissociation constant, = 1.85 × 10?5.  


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