Home
Class 12
CHEMISTRY
The resistance of 0.05M KCl solution at ...

The resistance of 0.05M KCl solution at `25^@C` is 100 ohm in a cell whose cell constant is `0.3765cm^(-1)`. The specific conductivity of KCl solution is

A

0.004765

B

0.003455

C

0.003765

D

0.003665

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • REVISION EXERCISE

    AAKASH SERIES|Exercise SURFACE CHEMISTRY|45 Videos

  • REVISION EXERCISE

    AAKASH SERIES|Exercise VA GROUP ELEMENTS|26 Videos

  • REVISION EXERCISE

    AAKASH SERIES|Exercise CHEMICAL KINETICS|39 Videos

  • REDOX REACTONS

    AAKASH SERIES|Exercise Questions For Descriptive Answers|27 Videos

  • SOLID STATE

    AAKASH SERIES|Exercise PRACTICE SHEET - 5|30 Videos

Similar Questions

Explore conceptually related problems

The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500Omega .What is the cell constant (in mm^(-1) ). If the conductivity of 0.001 M KCl solution is 2 xx 10^(-3) S mm^(-1)

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Omega . What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 xx 10^(-2) S cm^(-1) .

The resistance of 1N solution of acetic acid is 250 ohm when measured in a cell of cell constant 1.15cm^(-1) . The equivalent conductance ( in ohm^(-1)cm^(2)equi^(-1) ) of 1N acetic acid is

The resistance of 0.1 N solution of formic acid is 200 ohm and cell constant is 2.0cm^(-2) . The equivalent conductivity (in Scm^(2)eg^(-1) ) of 0.1 N formic acid is 1.0 xx 10^(x) , x is

The resistance of 0.1N KCl solution is found 702Omega awhen measured in a conductivity cell. The specific conductance of 0.1N KCl is 0.14807 Omega^(-1)m^(-1) . Calculate the cell constant.

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

The specific conductance of 0.1N KCl solution at M 23^@C is 0.012 ohm^(-1)cm^(-1) . The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be

0.05M NaOH solution offered a resistance of 31.69 Omega in a conductivity cell at 298K. If the cell constant of the conductivity cell is 0.367cm^(-1) the molar conductivity of the NaOH solution is

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100Omega . The conductivity of this solution is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.02M of the same solution is 520 Omega . The molar conductivity of 0.02M solution of the electrolyte will be

The conductivity of 0.20M solution of KCl at 298 K is 0.0248S cm^(-1). Calculate molar conductance.

AAKASH SERIES-REVISION EXERCISE -ELECTROCHEMISTRY
  1. On passing 3F electricity through three cells containing fused Na2CO3 ...

    Text Solution

    |

  2. For the electrolytic production of NaClO4 from NaClO3 according to t...

    Text Solution

    |

  3. The quantity of electricity in Faradays required to reduce 1.23 gm of ...

    Text Solution

    |

  4. When 3.86 amperes current are passed through an electrolyte for 50 min...

    Text Solution

    |

  5. When 965 amp current is passed through aqueous solution of salt X usin...

    Text Solution

    |

  6. What is the approximate quantity of electricity (in coulomb) required ...

    Text Solution

    |

  7. Four moles of electrons were transferred from anode to cathode in an e...

    Text Solution

    |

  8. One Faraday of electricity is passed separately through one litre of o...

    Text Solution

    |

  9. The resistance of 0.5 N solution of an electrolyte in a conductivity c...

    Text Solution

    |

  10. 0.05M NaOH solution offered a resistance of 31.69Omega in a conducti...

    Text Solution

    |

  11. The resistance of 0.05M KCl solution at 25^@C is 100 ohm in a cell wh...

    Text Solution

    |

  12. Aluminium displaces hydrogen from acids, but copper does not. A galvan...

    Text Solution

    |

  13. When E(Ag^(+) | Ag)^@ = 0.80 volt and E(Zn^(2+)|Zn)^@ = 0.76 volt, wh...

    Text Solution

    |

  14. Deduce from the following E^@ values of half cells, what combination o...

    Text Solution

    |

  15. The half cell reaction for the corrosion are , 2H^(+) + (1)/(2) O2 + ...

    Text Solution

    |

  16. A cell constructed by coupling a standard copper electrode and a stand...

    Text Solution

    |

  17. E^@ values of Zn^(2+), Zn and Cu^(2+), Cu are respectively -0.76 V a...

    Text Solution

    |

  18. E^@ of Fe|Fe^(2+) is +0.44V , E^@ of Cu|Cu^(2+) is -0.32V. Then in the...

    Text Solution

    |

  19. The standard reduction potentials of Ag, Cu, Co and Zn are 0.799, 0.33...

    Text Solution

    |

  20. The standard reduction potentials of Zn^(2+)//Zn,Cu^(2+)//Cu and Ag^(+...

    Text Solution

    |