Home
Class 12
CHEMISTRY
At 0.04 M concentration, the molar condu...

At 0.04 M concentration, the molar conductivity of solution of an electrolyte is `5000 Omega^(-1)cm^(2) mol^(-1)` while at 0.01 M concentration the value is `5100 Omega^(-1)cm^(2)mol^(-1)`. Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and write percentage dissociation of strong electrolyte at 0.04 M.

Text Solution

Verified by Experts

The correct Answer is:
96
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTY

    GRB PUBLICATION|Exercise Match the column type|16 Videos

  • D-BLOCK ELEMENTS

    GRB PUBLICATION|Exercise Subjective Type|18 Videos

  • ENVIRONMENTAL CHEMISTRY

    GRB PUBLICATION|Exercise Straight objective type|40 Videos

Similar Questions

Explore conceptually related problems

At 0.04 M concentration the molar conductivity of a solution of a electrolyte is 5000Omega^(-1)cm^(2)mol^(-1) while at 0.01 M concentration the value is 5100Omega^(-1)cm^(2)mol^(-1) making necessary assumption (taking it as strong electrolyte) find the molar conductivity at infinite dilution and also determine the degree of dissociation of strong electrolyte at 0.04M.

The molar conductivity of 0.05 M BaCl_(2) solution at 25^(@)C is 223 Omega^(-1)"cm"^(2) mol^(-1) . What is its conductivity?

The molar conductivity of 0.05 M solution of weak acid is 16.6Omega^(-1)cm^(-2)mol^(-1). Its molar conductivity at infinite dilution is 390.7Omega^(-1)cm^(-2)mol^(-1) . The degree of dissociation of weak acid is

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm^(2) mol^(-1) . Calculate the conductivity of this solution.

Molar conductivity of a 1.5 m solustion of an electrolyte is found to be 138.9 S cm^(2) mol^(-1) . Calculate the conductivity of this solution.

Molar conductivity of a solutions is 1.26xx10^(2) Omega^(-1)cm^(2) "mol"^(-1) its molarity is 0.01. its specific conductivity will be

The molar conductivity of 0.05 M of solution of an electrolyte is 200 Omega^(-1) cm^2 . The resistance offered by a conductivity cell with cell constant (1//3) cm^(-1) would be about .

GRB PUBLICATION-ELECTROCHEMISTY-Subjective type
  1. The conductivity ofa solution may be taken to be directly proportional...

    Text Solution

    |

  2. At 0.04 M concentration, the molar conductivity of solution of an elec...

    Text Solution

    |

  3. The EMF of the cell M|M^(n+) (0.02 M) | H^(+) (1 M) | H(2)(g) (1 atm),...

    Text Solution

    |

  4. A current of 0.5 ampere when passed through AgNO(3) solution for 193 s...

    Text Solution

    |

  5. A certain metal salt solution is electrolysed in series with a silver ...

    Text Solution

    |

  6. 3 Amp current was passed through an aquesous solution salt of unknown ...

    Text Solution

    |

  7. How long (in sec) a current of 2A has to be passed through a solution ...

    Text Solution

    |

  8. A fuell cell uses CH(4)(g) and forms CO(3)^(2-) at the anode. It is us...

    Text Solution

    |

  9. In a conductivity cell, the two platinum electrodes, each of area 10sq...

    Text Solution

    |

  10. Given the equivalent conductance of sodium butyrate, sodium chloride a...

    Text Solution

    |

  11. Calcualte the emf of the cell in mV (at least first two digits must ma...

    Text Solution

    |

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

    Text Solution

    |

  13. A silver coulometer is in series with a cell electrolyzing water. In a...

    Text Solution

    |

  14. After electrolytes of NaCl solution with inert electrodes for a certai...

    Text Solution

    |

  15. For the cells in opposition, Zn(s) | ZnCl(2)(sol).|AgCl(s)|Ag|AgCl(s...

    Text Solution

    |

  16. At Tl^(+) |Tl couple was prepared by saturating 0.1 M KBr with TlBr an...

    Text Solution

    |

  17. For the cell (at 1bar H(2) pressure) Pt|H(2)(g) HX(m(1), NaX(m(2), NaC...

    Text Solution

    |

  18. NO(3)^(-) rightarrow NO(2)(acid medium), E^(@0 = 0.790)V NO(3)^(-) r...

    Text Solution

    |

  19. EMF of the following cell is 0.634 volt at 298 K. ltbr. Pt|H92)(1 atm)...

    Text Solution

    |

  20. During electrolytes 40 mA current is passed through 100 mL of 0.2 M Fe...

    Text Solution

    |