Home
Class 12
CHEMISTRY
Calculate the equivalent conductivity of...

Calculate the equivalent conductivity of `1M H_(2)SO_(4)`, whose specific conductivity is `26 xx 10^(-2) "ohm"^(-1) cm^(-1).`

Text Solution

AI Generated Solution

To calculate the equivalent conductivity of \(1M \, H_2SO_4\) with a specific conductivity of \(26 \times 10^{-2} \, \text{ohm}^{-1} \, \text{cm}^{-1}\), we will follow these steps: ### Step 1: Understand the Given Data - Specific conductivity (\(k\)) = \(26 \times 10^{-2} \, \text{ohm}^{-1} \, \text{cm}^{-1}\) - Concentration of \(H_2SO_4\) = \(1M\) ### Step 2: Calculate the Molar Mass of \(H_2SO_4\) The molar mass of \(H_2SO_4\) is calculated as follows: ...
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    ICSE|Exercise ISC EXAMINATION QUESTIONS (PART-II Descriptive Questions) |11 Videos
  • DISTINCTION BETWEEN PAIRS OF COMPOUNDS

    ICSE|Exercise QUESTIONS |155 Videos
  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    ICSE|Exercise EXERCISE (PART-I ( OBJECTIVE QUESTIONS )) (Fill in the blanks)|60 Videos

Similar Questions

Explore conceptually related problems

Select the equivalent conductivity of 1.0 M H_(2)SO_(4) , if its conductivity is 0.26 ohm^(-1) cm^(-1) :

Select the equivalent conductivity of 1.0 M H_(2)SO_(4) , if its conductivity is 0.26 ohm^(-1) cm^(-1) :

Which of the following have maximum molar conductivity. (i) 0.08M solution and its specific conductivity is 2 xx 10^(-2) Omega^(-1) cm^(-1) (ii) 0.1M solution and its resistivity is 50 Omega cm .

Calculate the value of equivalent conductivity of MgCl_(2) at infinite dilution if lambda^(oo)(Mg^(2+))= 106.12 "ohm"^(-1)cm^(-2)"mol"^(-1), lambda^(oo)(Cl^(-))=76.34"ohm"^(-1)cm^(2)"mol"^(-1) .

Metals have conductivity in the order of (ohm^(-1) cm^(-1))

Metals have conductivity in the order of (ohm^(-1) cm^(-1))

Resistance of a conductivity cell filled with 0.1 M KCl is 100 ohm. If the resistence of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. Conductivity of 0.1 KCl solution is 1.29xx10^(-2)" ohm "^(-1)cm^(-1) .

The equivalent conductivity of H_(2)SO_(4) at infinite dilution is 384 Omega^(-1) cm^(2) eq^(-1) . If 49g H_(2)SO_(4) per litre is present in solution and specific resistance is 18.4 Omega then calculate the degree of dissociation.

The equivalent conductances at infinite dilution of NH_(4)Br, KOH and KBr" at "25^(@)C " are "271.52 xx 10^(-4) , 266.5 xx 10^(-4) and 1.51.66 xx 10^(-4)Omega^(-1)m^(2) ("g. eq")^(-1) respectively. Calculate the degree of dissociation of 0.01 N NH_(4)OH at 25^(@)C if the equivalent conductance of 0.01 N NH_(4)OH solution is 16.28 xx 10^(-4)Omega^(-1) m^(2) ("g eq")^(-1) .

The conductivity of 0*2M " KCl solution is " 3 Xx 10^(-2) ohm^(-1) cm^(-1) . Calculate its molar conductance.

ICSE-ELECTROCHEMISTRY-ISC EXAMINATION QUESTIONS (PART-II Numerical Problems)
  1. A cell is constructed by dipping a zinc rod in 0.1 M zinc nitrate solu...

    Text Solution

    |

  2. For the cell : Zn"|"Zn^(2+) (a=1)"||"Cu^(2+) (a=1)"|"Cu Given that ...

    Text Solution

    |

  3. Calculate the equivalent conductivity of 1M H(2)SO(4), whose specific...

    Text Solution

    |

  4. A current of 10 A is passed for 80 min and 27 seconds through a cell ...

    Text Solution

    |

  5. Calculate E("cell") at 25^(@)C for the reaction : Zn+Cu^(2+) (0.2...

    Text Solution

    |

  6. A 0.05 M NaOH solution offered a resistance of 31.6 ohms in a conduct...

    Text Solution

    |

  7. For the following cell, calculate the emf: Al//Al^(3+) (0.01 M)"||"...

    Text Solution

    |

  8. A solution of 0.1 N KCl offers a resistance of 245 ohms. Calculate th...

    Text Solution

    |

  9. How many electrons will flow when a current of 5 amperes is passed thr...

    Text Solution

    |

  10. Consider the reaction 2Ag^(+) +Cd to 2Ag to 2Ag +Cd^(2+). The stan...

    Text Solution

    |

  11. Calculate the maximum work that can be obtained from the given electr...

    Text Solution

    |

  12. Calculate the value of E("cell") at 298 K for the following cell: ...

    Text Solution

    |

  13. 0*05 " M " NaOH solution offered a resistance of 31*6 ohm in a conduct...

    Text Solution

    |

  14. 0*3605 g of a metal is deposited on the electrode by passing 1*2 amper...

    Text Solution

    |

  15. How many hours does it take to reduce 3 moles of Fe^(3+) to Fe^(2+) wi...

    Text Solution

    |

  16. Calculate the emf of the following cell reaction at 298 K: Mg(s) +C...

    Text Solution

    |

  17. The specific conductance of a 0.01 M solution of acetic acid at 298 K...

    Text Solution

    |

  18. Calculate the number of coulombs required to deposit 5.4g of Al when ...

    Text Solution

    |

  19. A 0.05 M NH OH solution offers the resistance of 50 ohms to a conduct...

    Text Solution

    |