Conductance — Λm, Kohlrausch's law
Electrochemistry
9
JEE Qs
8%
Hard
60
min
Master unit conversions and understand the qualitative differences in conductance behavior between strong and weak electrolytes with dilution.
🧮 Key Formulas
✅ Key Points for JEE
- 1Unit consistency is paramount in calculations; ensure κ (S cm⁻¹ or S m⁻¹), C (mol L⁻¹ or mol m⁻³), and Λm (S cm² mol⁻¹ or S m² mol⁻¹) are correctly interconverted.
- 2Specific conductance (κ) generally decreases with dilution for both strong and weak electrolytes, while molar conductivity (Λm) increases with dilution for both.
- 3Strong electrolytes show a linear decrease in Λm with sqrt(C) (Debye-Hückel-Onsager), whereas weak electrolytes show a very sharp increase in Λm at extreme dilution due to increased degree of dissociation.
- 4Kohlrausch's Law allows calculation of Λm^0 (limiting molar conductivity) for weak electrolytes by algebraically combining Λm^0 values of suitable strong electrolytes.
- 5The degree of dissociation (α) for a weak electrolyte can be determined from its molar conductivity at a given concentration and its limiting molar conductivity (α = Λm / Λm^0).
⚠️ Common Mistakes
- ✕Errors in unit conversions (e.g., L to cm³, S cm⁻¹ to S m⁻¹, mol L⁻¹ to mol m⁻³), leading to incorrect magnitude of Λm.
- ✕Confusing the trends of specific conductance (κ) and molar conductivity (Λm) with dilution for strong vs. weak electrolytes.
- ✕Incorrectly applying Kohlrausch's law for calculating Λm at finite concentration instead of limiting molar conductivity (Λm^0).
- ✕Misinterpreting the significance of the steep rise in Λm for weak electrolytes at high dilution compared to strong electrolytes.
📝 Practice Questions
See allQ54.The element that does not belong to the same period of the remaining elements (modern periodic table) is: 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) Iridium (2) Platinum (3) Osmium (4) Palladium
Q67.For the given cell Fe2+ (aq) + Ag+ (aq) →Fe3+(aq) + Ag(s) The standard cell potential of the above reaction is Ag+ + e−→Ag Eθ = xV Given: Fe2+ + 2e−→Fe Eθ = yV Fe3+ + 3e−→Fe Eθ = zV (1) x + y −z (2) x + 2y (3) x + 2y −3z (4) y −2x
Q58.Which of the following electrolyte can be used to obtain H2 S2O8 by the process of electrolysis? (1) Dilute solution of sodium sulphate. (2) Acidified dilute solution of sodium sulphate. (3) Dilute solution of sulphuric acid (4) Concentrated solution of sulphuric acid
Q66.A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is [Given : molar mass of aluminium and chlorine are 27 g mol−1 and 35.5 g mol−1 respectively. Faraday constant = 96500Cmol−1] (1) 1.660 g (2) 0.336 g (3) 0.441 g (4) 1.007 g 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
Q58.Standard electrode potentials for a few half cells are mentioned below : E∘ = 0.34 V, E∘ = −0.76 V Cu2+/Cu Zn2+/Zn Which one of the following cells gives the most negative value of E∘Ag+/Ag = 0.80 V, E∘Mg2+/Mg = −2.37 V ΔG∘ ? (1) Zn Zn2+(1M) Ag+(1M) Ag (2) Zn Zn2+(1M) Mg2+(1M) Mg (3) Ag Ag+(1M) Mg2+(1M) Mg (4) Cu Cu2+(1M)∥Ag+(1M) Ag
Q57. E∘Cr2O2−7 /Cr3+ = 1.33 V E∘Cl2/Cl(−) = 1.36 V Based on the data given below : the strongest reducing agent is : E∘ = 1.51 V E∘ = −0.74 V MnO−4 /Mn2+ Cr3+/Cr (1) Cr (2) Cl− (3) MnO−4 (4) Mn2+
NCERT Chapters
- Class 12 Chemistry Part 1 Ch 3: Electrochemistry