STANDARD Chemistry Molar Conductivity MCQ Question
Limiting molar conductivity of NaBr is
ΛₘNaBr = ΛₘNaCl + ΛₘKBr
ΛₘNaBr = ΛₘNaCl + ΛₘKBr − ΛₘKCl
ΛₘNaBr = ΛₘNaOH + ΛₘNaBr − ΛₘNaCl
ΛₘNaBr = ΛₘNaCl − ΛₘNaBr
Correct Answer
Detailed Explanation
The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. For NaBr, it is calculated as ΛₘNaCl + ΛₘKBr − ΛₘKCl.
Found an issue with this question?
Related Questions
More from Electrochemistry
E° value of Ni²⁺/Ni is^{-0}.25 V and Ag⁺/Ag is +0.80 V. If a cell is made by taking the two electrodes what is the feasibility of the reaction?
The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option. [Λ°_H⁺ =...
An electric charge of 5 Faradays is passed through three electrolytes AgNO₃, CuSO₄ and FeCl₃ solution. The grams of each metal liberated at cathode wi...
More from
Calculate the dissociation constant \( K_a \) for acetic acid given that its conductivity is \( 4.95 \times 10^{-5} \text{ S cm}^{-1} \) at a concentr...
Match Column-I with Column-II. Column-I Column-II (a) Cathode reaction (i) Cu 2+ (aq) + 2e – → Cu (s) (b) Anode reaction (ii) Cu(s) ...