Home
Class 12
CHEMISTRY
Equivalent conductivity of Fe(2)(SO(4))(...

Equivalent conductivity of `Fe_(2)(SO_(4))_(3)` is relative to molar conductivity by the expression `:`

A

`Lambda_(eq)=Lambda_(m)`

B

`Lambda_(eq)=lambda_(m)//3`

C

`Lambda_(eq)=3Lambda_(m)`

D

`Lambda_(eq)=Lambda_(m)//6`

Text Solution

AI Generated Solution

The correct Answer is:
To find the relationship between the equivalent conductivity and molar conductivity of \( \text{Fe}_2(\text{SO}_4)_3 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Definitions**: - **Molar Conductivity (\( \Lambda_m \))**: It is defined as the conductivity of an electrolyte solution divided by its molar concentration. It indicates how well an electrolyte conducts electricity in solution. - **Equivalent Conductivity (\( \Lambda_{eq} \))**: It is defined as the conductivity of an electrolyte solution divided by its equivalent concentration. It indicates the conductivity per equivalent of the solute. 2. **Determine the Number of Ions**: - The formula \( \text{Fe}_2(\text{SO}_4)_3 \) dissociates in solution to give: - 2 Fe\(^{3+}\) ions - 3 SO\(_4^{2-}\) ions - Therefore, the total number of ions produced from one formula unit of \( \text{Fe}_2(\text{SO}_4)_3 \) is \( 2 + 3 = 5 \) ions. 3. **Relating Molar Conductivity and Equivalent Conductivity**: - The molar conductivity can be expressed in terms of equivalent conductivity as follows: \[ \Lambda_m = n \cdot \Lambda_{eq} \] where \( n \) is the number of equivalents produced from one mole of the solute. 4. **Calculating the Number of Equivalents**: - For \( \text{Fe}_2(\text{SO}_4)_3 \), since it produces 5 ions, we can say that: \[ n = 5 \] 5. **Substituting into the Equation**: - Now substituting \( n \) into the relationship: \[ \Lambda_m = 5 \cdot \Lambda_{eq} \] 6. **Finding Equivalent Conductivity**: - Rearranging the equation to find \( \Lambda_{eq} \): \[ \Lambda_{eq} = \frac{\Lambda_m}{5} \] ### Final Expression: Thus, the equivalent conductivity of \( \text{Fe}_2(\text{SO}_4)_3 \) is related to its molar conductivity by the expression: \[ \Lambda_{eq} = \frac{\Lambda_m}{5} \]

To find the relationship between the equivalent conductivity and molar conductivity of \( \text{Fe}_2(\text{SO}_4)_3 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Definitions**: - **Molar Conductivity (\( \Lambda_m \))**: It is defined as the conductivity of an electrolyte solution divided by its molar concentration. It indicates how well an electrolyte conducts electricity in solution. - **Equivalent Conductivity (\( \Lambda_{eq} \))**: It is defined as the conductivity of an electrolyte solution divided by its equivalent concentration. It indicates the conductivity per equivalent of the solute. ...
Promotional Banner

Topper's Solved these Questions

  • DPP

    RESONANCE|Exercise QUESTIONS|223 Videos
  • ELECTROCHEMISRY

    RESONANCE|Exercise Advanced Level Problems|88 Videos

Similar Questions

Explore conceptually related problems

The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and 0.002768 S ch^(-1) respectively . When filled with 0.005 N K_(2)SO_(4) , the solution had a resistance of 324ohm . Calculate : a. Cell constant b. Conductance of K_(2)SO_(4) solution c. Conductivity of K_(2)SO_(4) solution d. Equivalent conductivity of K_(2)SO_(4) solution e. Molar conductivity of K_(20SO_(4) solution.

Equivalent conductivity|| Variation OF conductivity and molar conductivity with dilution

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

Specific conductivity of 0.01N H_(2)SO_(4) solution is 6 xx 10^(-3) S cm^(-1) . Its molar conductivity is

RESONANCE-ELECTRO CHEMISTRY-PHYSICAL CHEMITRY (ELECTROCHEMISTRY)
  1. For the half cell At pH=2. Electrod potential is :

    Text Solution

    |

  2. In the electrolysis of an aqueous potassium sulphate solution, the Ph ...

    Text Solution

    |

  3. How many electrons are there in one coulomb of electricity?

    Text Solution

    |

  4. Electrolysis can be used to determine atomic masses. A current of 0.55...

    Text Solution

    |

  5. Calculate the current ( in Ma) required to deposit 0.195 g of platinum...

    Text Solution

    |

  6. An aqueous solution containing 1M each of Au^(3+),Cu^(2+),Ag^(+),Li^(...

    Text Solution

    |

  7. Based on the following information arrange four metals A,B,C and D in ...

    Text Solution

    |

  8. The standard electrode potential for the following reaction is +1.33V...

    Text Solution

    |

  9. Ag|AgCl|Cl^(-)(C(2))||Cl^(-)(C(1))||AgCl|Ag for this cell DeltaG is ne...

    Text Solution

    |

  10. Resistance of a decimolar solution between two electrodes 0.02 meter a...

    Text Solution

    |

  11. Equivalent conductivity of Fe(2)(SO(4))(3) is relative to molar condu...

    Text Solution

    |

  12. The limiting equivalent conductivity of NaCl, KCl and KBr are 126.5,1...

    Text Solution

    |

  13. The resistance of 0.1N solution of formic acid is 200 ohm and cell con...

    Text Solution

    |

  14. Given that ( ohm^(-1) cm^(2)eq^(-1)), T=298 K {:(lambda(E)^(oo) f o...

    Text Solution

    |

  15. Na- amalgam is prepared by electrolysis of NaCl solution using liquid...

    Text Solution

    |

  16. Find the thickness of the electro silver if the surface area over whic...

    Text Solution

    |

  17. In the given figure the electrolytic cell contains 1L of an aqueous 1M...

    Text Solution

    |

  18. The equilibrium Cu^(. .)(aq)+Cu(s) hArr2Cu^(.) established at 20^(@)...

    Text Solution

    |

  19. What is the cell entropy change ( in J K ^(-1)) of the following cell...

    Text Solution

    |

  20. Calculate the value of Lambda(m) ^prop for SrCl(2) in water at 25^(@)...

    Text Solution

    |