Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve it, If the equivalent conductance of 1M benzenoic acid is 12.8 ohm-1cm2eq-1and if the conductance of benzoate ion and H+ions are 43 and 288.42 ohm-1cm2eq-1respectively. Its degree of dissociation is :

If the equivalent conductance of 1M benzenoic acid is 12.8 ohm-1 cm2 eq-1 and if the conductance of benzoate ion and H+ ions are 43 and 288.42 ohm-1 cm2 eq-1 respectively. Its degree of dissociation is :

  • Option 1)

    39%

  • Option 2)

    3.9%

  • Option 3)

    0.35%

  • Option 4)

    0.039%

 

Answers (1)

best_answer

As we have learnt,

 

Application of Kohlrausch's law -

Determination of Degree of dissociation of weak electrolytes.

- wherein

Degree of dissociation

\alpha = \frac{\Lambda m}{\Lambda_{m}^{0} }

 

 \\*\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{12.8}{42 + 288.42} = 0.0387 \\*\%\;dissociation = 100\times \alpha = 3.87\%

 


Option 1)

39%

Option 2)

3.9%

Option 3)

0.35%

Option 4)

0.039%

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question