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  • Please,please help me - Redox Reaction and Electrochemistry - JEE Main-5

Conductivity of 0.00241 M Acetic acid is 7.896 \times 1o^{-5} S\;cm^2. Calculate its molar conductivity and if \Lambda^o_{CH_3COOH}is 390.5 S cm2 mol-1. Calculate its dissociation constant.

  • Option 1)

    2.54\times 10^{-7}

  • Option 2)

    2.68\times 10^{-7}

  • Option 3)

    2.43\times 10^{-7}

  • Option 4)

    2.23\times 10^{-7}

 

Answers (1)

best_answer

As we have learnt,

 

Application of Kohlrausch's law -

Calculation of molar conductivities of weak electrolyte at infinite dilution.

-

 

\\*\Lambda = K\times \frac{1000}{M} = 32.76 \\* \alpha = \frac{\Lambda_v}{\Lambda_m} = \frac{32.76}{390.5} = 0.1\times 10^{-3} = 0.01\% \\*K_\alpha = \frac{c\alpha^2}{1-\alpha} = \frac{0.00241(0.01)^2}{0.99} = 2.43\times 10^{-7}

 


Option 1)

2.54\times 10^{-7}

Option 2)

2.68\times 10^{-7}

Option 3)

2.43\times 10^{-7}

Option 4)

2.23\times 10^{-7}

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Plabita

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