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Conductance questions mix three quantities that sound alike, conductance, conductivity and molar conductivity, and add a factor of 1000 that students drop under pressure. This post sorts out the definitions, works a cell constant problem, shows how Kohlrausch's law gives the limiting molar conductivity of a weak acid and uses it to find a dissociation constant.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "conductance in electrolytic solutions, molar conductivities and their variation with concentration" and "Kohlrausch's law and its applications". The JEE (Advanced) 2026 brochure lists "specific, equivalent and molar conductivity, Kohlrausch's law". Check the current documents for your exam.
What carries the current in a solution?
Ions do. Electrons flow in the wires and electrodes, but not through the solution. This sounds obvious, yet Garnett and Treagust (1992) found that many final-year school students believed current in a solution is carried by drifting electrons, and Sanger and Greenbowe (1997) found college students who thought electrons flow through the electrolyte. Keep the picture of moving ions in mind and the trends below make sense.
How are conductivity and molar conductivity defined?
The NCERT chapter Electrochemistry defines:
Conductance G = 1/R, in siemens (S).
Cell constant G* = l/A, which NCERT says is usually found by measuring the resistance of a KCl solution of known conductivity: G* = κR.
Conductivity κ = G*/R, in S cm⁻¹.
Molar conductivity Λm = κ/c. With κ in S cm⁻¹ and c in mol L⁻¹, Λm = κ × 1000/c in S cm² mol⁻¹. NCERT notes that 1 S cm² mol⁻¹ = 10⁻⁴ S m² mol⁻¹.
NCERT's Table 2.3 gives KCl at 298.15 K: 0.100 M has κ = 0.0129 S cm⁻¹ and Λm = 129.0 S cm² mol⁻¹, while 0.010 M has κ = 0.00141 S cm⁻¹ and Λm = 141.0. Conductivity falls on dilution, because there are fewer ions per unit volume. Molar conductivity rises, because the volume containing one mole increases faster.
Worked example 1: finding the cell constant
A conductivity cell filled with 0.100 M KCl has a resistance of 150 Ω. Filled with 0.050 M NaCl, the same cell reads 340 Ω. Find κ and Λm for the NaCl solution.
G* = κR = 0.0129 × 150 = 1.935 cm⁻¹.
κ(NaCl) = G*/R = 1.935/340 = 5.69 × 10⁻³ S cm⁻¹.
Λm = κ × 1000/c = 5.69 × 10⁻³ × 1000/0.050 = 114 S cm² mol⁻¹.
How does molar conductivity change with concentration?
For strong electrolytes, NCERT gives Λm = Λ°m − A√c, a straight line against √c. In NCERT Example 2.6, KCl data extrapolate to Λ°m = 150.0 S cm² mol⁻¹ with A = 87.46. As a check, at c = 0.010 M this gives 150.0 − 87.46 × 0.1 ≈ 141.3, close to the 141.0 in Table 2.3.
Weak electrolytes behave differently. Their Λm rises steeply at low concentration as more molecules dissociate, so you cannot extrapolate to find Λ°m. That is where Kohlrausch's law comes in.
What does Kohlrausch's law say?
NCERT describes how Kohlrausch noticed that Λ°m(KX) − Λ°m(NaX) is nearly the same, about 23.4 S cm² mol⁻¹, whatever the anion X. He concluded that the limiting molar conductivity of an electrolyte is the sum of independent contributions from its ions: Λ°m = ν₊λ°₊ + ν₋λ°₋.
NCERT's Table 2.4 gives, at 298 K, λ° values including H⁺ 349.6, K⁺ 73.5, Cl⁻ 76.3, CH₃COO⁻ 40.9 and SO₄²⁻ 160.0 S cm² mol⁻¹. So Λ°m(K₂SO₄) = 2(73.5) + 160.0 = 307.0 S cm² mol⁻¹. The coefficient 2 is easy to forget.
Worked example 2: dissociation constant of acetic acid
Suppose 0.010 M acetic acid has κ = 1.65 × 10⁻⁴ S cm⁻¹. Find α and Ka.
Λm = 1.65 × 10⁻⁴ × 1000/0.010 = 16.5 S cm² mol⁻¹.
α = Λm/Λ°m = 16.5/390.5 = 0.0423.
Ka = cα²/(1 − α) = 0.010 × (0.0423)²/(0.958) ≈ 1.9 × 10⁻⁵.
NCERT Example 2.9 uses the same steps for a 0.001028 M solution and gets 1.78 × 10⁻⁵. The two results are close, which is a useful check on your arithmetic.
Which mistakes cost marks?
Dropping the 1000 when c is in mol L⁻¹.
Saying conductivity increases on dilution. Molar conductivity does; conductivity does not.
Trying to extrapolate a weak electrolyte's curve to zero concentration.
Forgetting ion coefficients in Kohlrausch sums, as in K₂SO₄ or CaCl₂.
Mixing S cm² mol⁻¹ with S m² mol⁻¹.
A practice routine
Write the four definitions and their units from memory.
Solve NCERT in-text question 2.9 on methanoic acid using the λ° values given there.
Build three Kohlrausch sums for weak acids or bases from strong-electrolyte data.
Connect α to pH: once you have Ka, find the pH of the same solution. Our guide to ionic equilibrium and pH shows how.
For the other half of the chapter, read our post on electrolysis and Faraday's laws. If a definition still feels slippery, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise conductivity on its own.
This week, redo worked example 2 with κ = 5.0 × 10⁻⁴ S cm⁻¹ at 0.10 M and compare the Ka you get.
