A conductance cell filled with 0.01 M solution of an electrolyte offered a resistance of 2300 ohms. If the electrodes of the cell are 1.90 cm apart and have an area of cross-section equal to 4.70 sq cm, calculate specific conductance

A conductance cell filled with 0.01 M solution of an electrolyte offered a resistance of 2300 ohms. If the electrodes of the cell are 1.90 cm apart and have an area of cross-section equal to 4.70 sq cm, calculate specific conductance

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter22: Inorganic Materials

Section: Chapter Questions

Problem 22P

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The following cell was found to have a potential of 0.124 V: Ag|AgCl(sat’d)||Cu2+ (3.25 10-3 M|mernbrane electrode for Cu2+ When the solution of known copper activity was replaced with an unknown solution, the potential was found to be 0.055 V. What was the pCu of this unknown solution? Neglect the junction potential.
.A conductance cell filled with 0.01 M solution of an electrolyte offered a resistance of 2300 ohms. If the electrodes of the cell are 1.90 cm apart and have an area of cross-section equal to 4.70 sq cm, calculate specific conductance Calculate the Molar conductivity
.A conductance cell filled with 0.01 M solution of an electrolyte offered a resistance of 2300 ohms. If the electrodes of the cell are 1.90 cm apart and have an area of cross-section equal to 4.70 sq cm, calculate specific conductance Calculate molar activity
The boundary equivalent conductivity of the 0.1 M electrolyte with a resistance of 0.33 ohms was determined as 320 Scm-2mol-1 and the conductivity ratio was determined as 0.55. According to this information, the electrolyte; a) Equivalent conductivity b) Self conductivity c) Find the cell constant of the electrode used.
The boundary equivalent conductivity of the 0.1 M electrolyte with a resistance of 0.33 ohms was determined as 320 Scm-2mol-1 and the conductivity ratio was determined as 0.55. According to this information, the electrolyte; a) Equivalent conductivityb) Self conductivityb) Find the cell constant of the electrode used.