The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm² mol⁻¹ respectively. The molar conductivity at infinite dilution of barium sulphate is _________ S cm² mol⁻¹. (Round off to the Nearest Integer ).

Solution

Chemistry ma’am · 4 steps

Add BaCl₂ and H₂SO₄, then take away two HCl to cancel the extra ions.

1. Kohlrausch's law. At infinite dilution every ion conducts independently. So a compound's Λₘ° is the sum of its ions' contributions, each counted as many times as the ion appears. We want λ°(Ba²⁺) + λ°(SO₄²⁻).
2. Pick the combination. BaCl₂ supplies Ba²⁺ along with 2Cl⁻, and H₂SO₄ supplies SO₄²⁻ along with 2H⁺. Removing 2H⁺ and 2Cl⁻ means subtracting two units of HCl.

   Λₘ°(BaSO₄) = Λₘ°(BaCl₂) + Λₘ°(H₂SO₄) - 2Λₘ°(HCl)

3. Put in numbers. 280 + 860 - 2 × 426 = 1140 - 852 = 288 S cm² mol⁻¹.
4. So. The molar conductivity of BaSO₄ at infinite dilution is 288 S cm² mol⁻¹.

The common slip: 714. Subtracting Λₘ°(HCl) only once gives 280 + 860 - 426 = 714. Two HCl are needed to cancel 2H⁺ and 2Cl⁻.

The idea to keep. For Kohlrausch questions, write each compound as its ions and choose the multipliers so the unwanted ions cancel exactly. Always check the ion counts on both sides.

Answer from the NTA final answer key. Solution by PrepWiser.