Home
Class 12
CHEMISTRY
The resistance of 0.01N solution of an e...

The resistance of `0.01N` solution of an electrolysis is `210 Omega` at `298 K` with a cell constant of `0.88 cm^(-1)`. Calculate the conductivity and equivalent conducitivty of the solution.

Text Solution

AI Generated Solution

To solve the problem, we need to calculate the conductivity (K) and equivalent conductivity (λ) of a 0.01 N solution of an electrolyte given its resistance and cell constant. ### Step-by-Step Solution: **Step 1: Calculate Conductivity (K)** The formula for conductivity (K) is given by: ...
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples (Electrochemical Cell)|36 Videos

  • ELECTROCHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples(Electrolysis And Electrolytic Cells)|12 Videos

  • D AND F BLOCK ELEMENTS

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|29 Videos

  • GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives (Subjective)|14 Videos

Similar Questions

Explore conceptually related problems

The resistance of a 0.01N solution of an electrolyte was found to 210 ohm at 298K using a conductivity cell with a cell constant of 0.88 cm^(-1) . Calculate specific conductance and equivalent conductance of solution.

The resistance of a 0.01N solution of an electroyte was found to 210 ohm at 298K using a conductivity cell with a cell constant of 0.88 cm^(-1) . Calculate specific conductance and equilvalent conductance of solution.

The resistance of 0.01 N solution of an electrolyte waw found to be 210 ohm at 298 K, when a conductivity cell with cell constant 0.66 cm^(-1) is used. The equivalent conductance of solution is:

0.05 "M" NaOH solution offered a resistance of 31.6 ohm in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^(_1) calculate the molar conductivity of the NaOH solution.

(a) What are the two classifications of batteries ? What is the difference between them ? (b) The resistanc of 0.01 M NaCl solution at 25^(@)C is 200Omega. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution.

The resistance of 0.01N solution of an electrolyte AB at 328K is 100ohm. The specific conductance of solution is (cell constant = 1cm^(-1) )

The resistance of 0.01 M NaCl solution at 25°C is 200 ohm.The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.

The resistance of 0.1N solution of formic acid is 200 ohm and cell constant is 2.0 cm^(-1) . The equivalent conductivity ( in Scm^(2) eq^(-1)) of 0.1N formic acid is :

(a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride. (b) Given are the conductivity and molar conductivity of NaCI solutions at 298 K at different concentrations : Compare the variation of conductivity and molar conductivity of NaCI solutions on dilution. Give reason. (c) 0.1 M KCI solution offered a resistance of 100 ohms in conductivity cell at 298 K. If the cell constant of the cell is 1.29 cm^(-1) , calculate the molar conductivity of KCI solution.

The resistance of 0.01 N solution at 25^(@)C is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

CENGAGE CHEMISTRY ENGLISH-ELECTROCHEMISTRY-Archieves Subjective
  1. The resistance of 0.01N solution of an electrolysis is 210 Omega at 29...

    Text Solution

    |

  2. A current of 3.7 A is passed for 6hrs. Between Ni electrodes in 0.5 L ...

    Text Solution

    |

  3. Consider the cell : Zn|Zn^(2+)(aq)(1.0M)||Cu^(2+)(aq)(1.0M)||Cu Th...

    Text Solution

    |

  4. In an electrolysis experiment, current was passed for 5h through two c...

    Text Solution

    |

  5. How long a current of 3A has to be passed through a solution of silver...

    Text Solution

    |

  6. Give reasons in one or two sentences :" anhydrous HCl is a bad conduc...

    Text Solution

    |

  7. The EMF of the following cellis 1.05V at 25^(@)C: Pt,H(2)(g)(1.0 atm...

    Text Solution

    |

  8. During the discharge of a lead storage battery, the density of sulphur...

    Text Solution

    |

  9. A 100 watt, 110 volt incandescent lamp is connected in series with an ...

    Text Solution

    |

  10. A cell contains two hydrogen electrodes. The negative electrode is in ...

    Text Solution

    |

  11. In a fuel cell, hydrogen and oxygen react to produce electricity. In p...

    Text Solution

    |

  12. An acidic solution of Cu^(2+0 salt containing 0.4g of Cu^(2+) is elect...

    Text Solution

    |

  13. Define Photolytic decomposition reaction.

    Text Solution

    |

  14. The standard reduction potential of Cu^(2+)|Cu and Ag^(o+)|Ag electrod...

    Text Solution

    |

  15. Calculate the quantity of electricity that would be required to reduce...

    Text Solution

    |

  16. Zinc granules are added in excess to 500mL OF 1.0m nickel nitrate solu...

    Text Solution

    |

  17. During the electrolysis of an aqueoius nitric acid solution using pt e...

    Text Solution

    |

  18. The standard reduction potential of Ag^(+)//Ag electrode at 298 K is 0...

    Text Solution

    |

  19. An aqueous solution of NaCl on electrolysis gives H(2)(g), Cl(2)(g), a...

    Text Solution

    |

  20. The standard reduction potential for the half cell : NO(3)^(c-)(aq)+...

    Text Solution

    |

  21. Chromium metal is electroplated using an acidic solution containing Cr...

    Text Solution

    |