Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Give answer! - Redox Reaction and Electrochemistry - JEE Main-2

A 0.01 M solution of KCl has a specific conductivity of 3.8\times 10^{-3} ohm^{-1}cm^{-1}. The resistance of the solution is 760\;ohm. If the resistance of a 0.01 M NaCl solution at 25o C (same cell constant) was 540 ohm. Calculate the molar conductivity of NaCl solution.

  • Option 1)

    535 \Omega cm^{2}mol^{-1}

  • Option 2)

    585 \Omega cm^{2}mol^{-1}

  • Option 3)

    555 \Omega cm^{2}mol^{-1}

  • Option 4)

    525 \Omega cm^{2}mol^{-1}

 

Answers (1)

best_answer

As we have learnt,

 

Molar Conductivity -

\Lambda m = \frac{K}{C}

- wherein

K is expressed in Sm^{-1} and concentration in mol m^{-3} .

Unit of \Lambda m = Sm^{2} mol^{-1}

 

 \frac{l}{A} = \frac{3.8\times 10^{-3}}{\frac{1}{760}} = 3.8\times 10^{-3}\times 760

Specific conductivity = cell constant x conductance = 3.8\times 10^{-3}\times 760\times \frac{1}{540}

Molar conductance = specific conductance x \frac{1000}{C}

                                =\frac{3.8\times 10^{-3}\times 760}{540} \times \frac{1000}{0.01} = 534.8 \approx 535

 


Option 1)

535 \Omega cm^{2}mol^{-1}

Option 2)

585 \Omega cm^{2}mol^{-1}

Option 3)

555 \Omega cm^{2}mol^{-1}

Option 4)

525 \Omega cm^{2}mol^{-1}

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 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question