ElectrochemistryhardNUMERICAL

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acElectrochemistry Chemistry Question

Question

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm mol respectively. The molar conductivity at infinite dilution of barium sulphate is ................... S cm mol . (Round off to the Nearest Integer). 2 –1 2 –1

Answer: 288.00

💡 Solution & Explanation

**Step 1: Identify the relevant ions** BaCl₂ provides Ba²⁺ and Cl⁻ ions H₂SO₄ provides H⁺ and SO₄²⁻ ions HCl provides H⁺ and Cl⁻ ions BaSO₄ would contain Ba²⁺ and SO₄²⁻ ions **Step 2: Apply Kohlrausch's Law of Independent Migration of Ions** Molar conductivity at infinite dilution equals the sum of ionic conductivities: Λ°(BaCl₂) = Λ°(Ba²⁺) + 2Λ°(Cl⁻) = 280 ... (1) Λ°(H₂SO₄) = 2Λ°(H⁺) + Λ°(SO₄²⁻) = 860 ... (2) Λ°(HCl) = Λ°(H⁺) + Λ°(Cl⁻) = 426 ... (3) **Step 3: Find Λ°(Ba²⁺) + Λ°(SO₄²⁻)** From equation (2): Λ°(H⁺) + Λ°(SO₄²⁻) = 430 From equation (3): Λ°(H⁺) + Λ°(Cl⁻) = 426 Subtracting (3) from (2): Λ°(SO₄²⁻) - Λ°(Cl⁻) = 4 **Step 4: Calculate Λ°(BaSO₄)** Λ°(BaSO₄) = Λ°(Ba²⁺) + Λ°(SO₄²⁻) From (1): Λ°(Ba²⁺) = 280 - 2Λ°(Cl⁻) From (3): Λ°

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