Calculate molar conductivities at zero concentration for CaCl$_2$ and Na$_2$SO$_4$. Given : molar ionic conductivitis of Ca$^{2+}$, Cl$^-$, Na$^+$ and SO$_4^{2-}$ ions are respectively, 104, 76.4, 50.1 and 159.6 $\Omega^{-1}$ cm$^2$ mol$^{-1}$.

Calculate molar conductivities at zero concentration for CaCl₂ and Na₂SO₄. Given : molar ionic conductivitis o

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$\Lambda_0 = \sum \nu\,\lambda^0$ : CaCl $_2$ $\rightarrow$ 104 + 2(76.4) = 256.8; Na $_2$ SO $_4$ $\rightarrow$ 2(50.1) + 159.6 = 259.8 (both $\Omega^{-1}$ cm $^2$ mol $^{-1}$ ).

Step 1. According to Kohlrausch's law, the molar conductivity of an electrolyte at zero concentration is the sum of the molar ionic conductivities of its ions, each multiplied by the number of ions per formula unit.

Step 2. For CaCl $_2$ (one Ca $^{2+}$ , two Cl $^-$ ):

\Lambda_0(\mathrm{CaCl_2}) = \lambda^0_{Ca^{2+}} + 2\,\lambda^0_{Cl^-} = 104 + 2 \times 76.4 = 256.8\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}

Step 3. For Na $_2$ SO $_4$ (two Na $^+$ , one SO $_4^{2-}$ ): …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM43 marks
$Q. i) What is fuel cell? ii) At 298 K the molar conductances of KCl and LiCl at infinite dilution are Λ°M(KCl) = 149.9 and Λ°M(LiCl) = 115.0 ohm^-1 cm^2 mol^-1. Calculate the value of [Λ°M(KNO3) - Λ°M(LiNO3)].$
›Reveal solutionSolution
A fuel cell continuously turns a fuel + oxidant into electricity; and by Kohlrausch's law of independent ionic migration the common nitrate ion cancels, so Lambda(KNO3) - Lambda(LiNO3) equals Lambda(KCl) - Lambda(LiCl) = 34.9 ohm^-1 cm^2 mol^-1.

Fuel cell A fuel cell is a galvanic (voltaic) cell in which the chemical energy released by the combustion-type reaction of a fuel (such as hydrogen, methane or methanol) with an oxidant (oxygen) is converted directly and continuously into electrical energy, as long as the reactants are fed in from outside. In the H2-O2 fuel cell, at the anode H2 + 2OH- -> 2H2O + 2e- and at the cathode O2 + 2H2O + 4e- -> 4OH-, giving an overall reaction 2H2 + O2 -> 2H2O. It is efficient and pollution-free.

Using Kohlrausch's law of independent migration of ions At infinite dilution each ion contributes a fixed molar ionic conductance: Lambda(KNO3) = lambda(K+) + lambda(NO3-) Lambda(LiNO3) = lambda(Li+) + lambda(NO3-) Subtracting, the common lambda(NO3-) cancels: Lambda(KNO3) - Lambda(LiNO3) = lambda(K+) - lambda(Li+) The same difference lambda(K+) - lambda(Li+) is obtained from the chlorides (the common lambda(Cl-) cancels there too): lambda(K+) - lambda(Li+) = Lambda(KCl) - Lambda(LiCl) = 149.9 - 115.0 = 34.9 Therefore Lambda(KNO3) - Lambda(LiNO3) = 34.9 ohm^-1 cm^2 mol^-1. …
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CBSE 2023Set ANNUAL3 marks
$Q. (i) State the Kohlrausch's law of independent migration of ions.$
(ii) The specific conductance or conductivity of a 0.01 M acetic acid at 298 K is 1.65x10-4 S cm-1. Calculate molar conductivity of the solution and degree of dissociation of CH3COOH. Given that, lambda-degree(H+) = 349.1 and lambda-degree(CH3COO-) = 40.9 S cm2 mol-1. (1+2) OR
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(ii) Write Nernst equation of the following galvanic cell and calculate emf of the cell at 298 K temperature: Cu(s) | Cu2+(0.130M) || Ag+(1.0x10-4M) | Ag(s). Given that E-degree(Cu2+/Cu) = +0.34V, E-degree(Ag+/Ag) = +0.80V. (1+2)
›Reveal solutionSolution
Kohlrausch's law lets ionic contributions to conductivity be added independently; applying it to the given data gives a degree of dissociation of about 4.2% for this dilute acetic acid solution.

Kohlrausch's law of independent migration of ions states that at infinite dilution, each ion migrates independently of its co-ion, and the limiting molar conductivity of the electrolyte is the sum of the limiting ionic conductivities of the cation and anion: $\Lambda_m^{\circ} = u_+\lambda_+^{\circ} + u_-\lambda_-^{\circ}$

Molar conductivity from specific conductance: $\Lambda_m = \kappa \times \frac{1000}{C} = 1.65\times10^{-4} \times \frac{1000}{0.01} = 16.5\ \text{S cm}^2\text{mol}^{-1}$ Limiting molar conductivity via Kohlrausch's law: …
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Q.Calculate molar conductivities at zero concentration for CaCl $_2$ and Na $_2$ SO $_4$ . Given : molar ionic conductivitis of Ca $^{2+}$ , Cl $^-$ , Na $^+$ and SO $_4^{2-}$ ions are respectively, 104, 76.4, 50.1 and 159.6 $\Omega^{-1}$ cm $^2$ mol $^{-1}$ .

Concept understanding — Kohlrausch's Law

Q.Calculate molar conductivities at zero concentration for CaCl2_22​ and Na2_22​SO4_44​. Given : molar ionic conductivitis of Ca2+^{2+}2+, Cl−^-−, Na+^++ and SO42−_4^{2-}42−​ ions are respectively, 104, 76.4, 50.1 and 159.6 Ω−1\Omega^{-1}Ω−1 cm2^22 mol−1^{-1}−1.

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Q.Calculate molar conductivities at zero concentration for CaCl $_2$ and Na $_2$ SO $_4$ . Given : molar ionic conductivitis of Ca $^{2+}$ , Cl $^-$ , Na $^+$ and SO $_4^{2-}$ ions are respectively, 104, 76.4, 50.1 and 159.6 $\Omega^{-1}$ cm $^2$ mol $^{-1}$ .

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