Kohlrausch Law

Answer: A. Kohlrausch's law of independent migration of ions states that at infinite dilution, the contribution of each ion to the total molar conductivity is a constant, independent of the counterion (NCERT Class 12 Chemistry Chapter 2, page 50).

Why B is wrong: B is wrong because Kohlrausch's law is about independent ionic contributions at infinite dilution, not proportionality to concentration. At infinite dilution, concentration is effectively zero.

Why C is wrong: C is wrong because Kohlrausch's law applies to all electrolytes — strong and weak — at infinite dilution. Its practical utility for weak electrolytes (calculating Λ°_m indirectly) does not limit the law's scope.

Why D is wrong: D is wrong because Kohlrausch's law holds at any temperature. The λ° values themselves are temperature-dependent, but the additive principle is universal.

Answer: C. BaCl₂ dissociates into 1 Ba²⁺ and 2 Cl⁻ ions. By Kohlrausch's law: Λ°_m = ν₊λ°₊ + ν₋λ°₋ = 1 × λ°(Ba²⁺) + 2 × λ°(Cl⁻) (NCERT Class 12 Chemistry Chapter 2, page 50).

Why A is wrong: A is wrong because it ignores the stoichiometric coefficient for Cl⁻. BaCl₂ produces 2 Cl⁻ ions per formula unit, so the chloride contribution must be multiplied by 2.

Why B is wrong: B is wrong because Ba²⁺ has ν₊ = 1 (one barium ion per formula unit), not 2. The coefficient 2 belongs to Cl⁻.

Why D is wrong: D is wrong because neither ion has a coefficient of 2 for Ba²⁺. Only Cl⁻ has ν = 2 in BaCl₂.

Answer: B. Weak electrolytes do not reach a conductivity plateau at experimentally achievable dilutions, so Λ°_m cannot be measured directly. Kohlrausch's law provides an indirect calculation route using strong electrolyte data (NCERT Class 12 Chemistry Chapter 2, page 50).

Why A is wrong: A is wrong because cell EMF is calculated from standard reduction potentials using E°_cell = E°_cathode − E°_anode, which is a separate electrochemistry concept unrelated to Kohlrausch's law.

Why C is wrong: C is wrong because mass deposited during electrolysis is governed by Faraday's law (m = MIt/nF), not Kohlrausch's law.

Why D is wrong: D is wrong because spontaneity prediction uses ΔG° = −nFE° or the sign of E°_cell. Kohlrausch's law addresses conductivity, not thermodynamic spontaneity.

Answer: C. Λ°_m(KCl) = Λ°_m(KNO₃) + Λ°_m(NaCl) − Λ°_m(NaNO₃) = 145.0 + 126.4 − 121.6 = 149.8 S cm² mol⁻¹. The Na⁺ and NO₃⁻ contributions cancel, leaving K⁺ + Cl⁻ (NCERT Class 12 Chemistry Chapter 2, page 50).

Why A is wrong: A is wrong — this is the value of Λ°_m(NaNO₃) itself, not KCl. A student picking this likely confused which electrolyte was being asked for.

Why B is wrong: B is wrong — this is Λ°_m(KNO₃), not KCl. The student may have assumed KCl and KNO₃ have the same Λ°_m because they share K⁺, forgetting the anion contribution differs.

Why D is wrong: D is wrong — this is the sum of all three given values (126.4 + 121.6 + 145.0 = 393.0). The correct method requires addition and subtraction to cancel auxiliary ions, not blind summation.

Answer: B. Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl) = 91.0 + 426.2 − 126.4 = 390.8 S cm² mol⁻¹. Na⁺ and Cl⁻ cancel, yielding λ°(CH₃COO⁻) + λ°(H⁺) (NCERT Class 12 Chemistry Chapter 2, page 50).

Why A is wrong: A is wrong — 517.2 = 91.0 + 426.2. This results from forgetting to subtract Λ°_m(NaCl). Without the subtraction, Na⁺ and Cl⁻ are not cancelled and the answer double-counts ions.

Why C is wrong: C is wrong — 643.6 = 91.0 + 426.2 + 126.4. Adding all three values ignores the cancellation logic entirely. Kohlrausch's indirect method requires subtraction of the auxiliary electrolyte.

Why D is wrong: D is wrong — this value does not correspond to any valid combination. The student may have attempted subtraction in the wrong order or misidentified which electrolytes to combine.

Answer: A. Λ_m = 1000κ/C = 1000 × 1.25 × 10⁻⁴ / 0.025 = 0.125 / 0.025 = 5.0 S cm² mol⁻¹. The 1000 factor converts L to cm³ (NCERT Class 12 Chemistry Chapter 2, page 50).

Why B is wrong: B is wrong — 0.5 results from omitting the 1000 factor: (1.25 × 10⁻⁴)/0.025 × 100 = 0.5. The correct formula requires multiplication by 1000, not 100. This is a common unit-conversion distractor.

Why C is wrong: C is wrong — 50.0 results from multiplying by 10000 instead of 1000, likely from confusing the conversion factor between different unit systems (S m⁻¹ vs S cm⁻¹).

Why D is wrong: D is wrong — 3.125 × 10⁻⁶ = κ × C = 1.25 × 10⁻⁴ × 0.025. This inverts the formula, using Λ_m = κ × C instead of Λ_m = 1000κ/C. The NEET pattern: molar conductivity problem pattern confirms this inversion as a frequent NEET distractor.

Answer: D. Weak electrolytes like acetic acid show a sharp, unbounded rise in Λ_m as concentration approaches zero because the degree of dissociation increases rapidly. No experimentally accessible concentration is dilute enough to give the plateau that strong electrolytes reach (NCERT Class 12 Chemistry Chapter 2, page 50). This is precisely why Kohlrausch's indirect method is needed.

Why A is wrong: A is wrong — acetic acid is thermally and chemically stable in dilute aqueous solution. It does not decompose upon dilution.

Why B is wrong: B is wrong — Kohlrausch's law is universally valid at infinite dilution for all electrolytes, organic or inorganic. In fact, it is the standard method for obtaining Λ°_m of organic weak acids like acetic acid.

Why C is wrong: C is wrong — acetic acid is a weak electrolyte, not a non-electrolyte. It partially dissociates into CH₃COO⁻ and H⁺ in water and does conduct electricity.

Answer: B. B is correct. Kohlrausch's law lets the limiting molar conductivity of an electrolyte be built from those of others, because each ion contributes independently. Λ°_m(NH₄OH) = λ°(NH₄⁺) + λ°(OH⁻). Combine: Λ°_m(NH₄Cl) + ½Λ°_m(Ba(OH)₂) − ½Λ°_m(BaCl₂). The ½ Ba²⁺ terms cancel, the Cl⁻ from NH₄Cl cancels against the Cl⁻ removed with BaCl₂, and what is left is NH₄⁺ + OH⁻: 149.8 + 228.8 − 140.0 = 238.6 S cm² mol⁻¹ (NCERT Class 12 Chemistry Chapter 2, page 50).

Why A is wrong: A is wrong because 518.6 comes from ADDING the half of Λ°_m(BaCl₂) instead of subtracting it: 149.8 + 228.8 + 140.0. The BaCl₂ term is there to remove the unwanted Ba²⁺ and to cancel the Cl⁻ brought in by NH₄Cl, so it must be subtracted.

Why C is wrong: C is wrong because 61.0 comes from swapping the roles of the two barium salts: 149.8 + ½(280.0) − ½(457.6). That combination builds NH₄⁺ with two extra chlorides removed, not with hydroxide added.

Why D is wrong: D is wrong because 327.4 comes from leaving out the ½: 149.8 + 457.6 − 280.0. Each barium salt supplies TWO anions per formula unit, so only half of its molar conductivity corresponds to one OH⁻ or one Cl⁻.