JEE Main · 2024 · Shift-ImediumEC-096

Molar ionic conductivities of divalent cation and anion are 57\ S\ cm2\ mol-1 and 73\ S\ cm2\ mol-1 respectively. The…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

Molar ionic conductivities of divalent cation and anion are 57 S cm2 mol157\ \mathrm{S\ cm^2\ mol^{-1}} and 73 S cm2 mol173\ \mathrm{S\ cm^2\ mol^{-1}} respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:

Options
  1. a

    187 S cm2 mol1187\ \mathrm{S\ cm^2\ mol^{-1}}

  2. b

    260 S cm2 mol1260\ \mathrm{S\ cm^2\ mol^{-1}}

  3. c

    130 S cm2 mol1130\ \mathrm{S\ cm^2\ mol^{-1}}

  4. d

    65 S cm2 mol165\ \mathrm{S\ cm^2\ mol^{-1}}

Correct Answerc

130 S cm2 mol1130\ \mathrm{S\ cm^2\ mol^{-1}}

Detailed Solution

Strategy: Apply Kohlrausch's Law of independent migration of ions. The molar conductivity of an electrolyte is the sum of the products of molar ionic conductivities and their respective stoichiometric coefficients.

Step 1: Determine the formula An electrolyte with a divalent cation (\ceM2+\ce{M^{2+}}) and a divalent anion (\ceX2\ce{X^{2-}}) has the formula \ceMX\ce{MX}. Stoichiometric coefficients: u+=1u_+ = 1, u=1u_- = 1.

Step 2: Calculate Λm\Lambda_m

u_+ \lambda^\circ_+ + u_- \lambda^\circ_-$$ $$\Lambda_m = 1(57) + 1(73) = 130 \mathrm{S cm^2 mol^{-1}}$$ $$\boxed{\text{Answer: (c) } 130 \mathrm{S\ cm^2\ mol^{-1}}}$$

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