The pH and conductance of a weak acid (HX) was found to be 5 and 4 times 10^-5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1text cm^2 were at a distance of 15text cm apart. The value of the limiting molar conductivity is ..... textS m^2text mol^-1. (nearest integer) (Given: degree of dissociation of the weak acid (alpha) ll 1)

Numerical Answer Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

### Related Formula kappa = G times fraclA Lambda_m = frackappa times 1000C alpha = fracLambda_mLambda_m^infty [H^+] = Calpha ### Core Logic Given: pH = 5, so [H^+] = 10^-5text M. Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty, we have 10^-5 = C cdot fracLambda_mLambda_m^infty. First, calculate conductivity (kappa): Conductance G = 4 times 10^-5text S. Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1. kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1 Molar conductivity (Lambda_m): Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C Substitute Lambda_m into the proton concentration formula: [H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty 10^-5 = frac0.6Lambda_m^infty Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1 Convert units to textS m^2text mol^-1: Since 1text m^2 = 10^4text cm^2, Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Reference Study Guides

Electrochemistry

More Electrochemistry Previous-Year Questions — Page 6

Q65 jee_main_2024_31_jan_morning Batteries
The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:
  • A. textB, C and E only
  • B. textA, B, C, D and E
  • C. textA, B, C and D only
  • D. textB, D and E only

Solution

### Core Logic Mn, Ni, and Cd metals are predominantly used in battery industries. - Mn is used in dry cells (Leclanche cell). - Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q67 jee_main_2024_31_jan_morning Electrolytic Conductance
Identify the factor from the following that does not affect electrolytic conductance of a solution.
  • A. textThe nature of the electrolyte added.
  • B. textThe nature of the electrode used.
  • C. textConcentration of the electrolyte.
  • D. textThe nature of solvent used.

Solution

### Core Logic Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q90 jee_main_2024_31_jan_morning Faraday's Laws of Electrolysis
One Faraday of electricity liberates x times 10^-1 gram atom of copper from copper sulphate, x is
Numerical Answer. Answer: 5 to 5

Solution

### Core Logic The reduction reaction for copper is: Cu^2+ + 2e^- rightarrow Cu This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu. Therefore, 1 Faraday of electricity will deposit: frac12 = 0.5 text moles of Cu ### Step 1: Finding x 0.5 text mole = 0.5 text gram atom = 5 times 10^-1 text gram atom Hence, x = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

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