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Electrochemistry question

2025 · 3 Apr · Shift 1 · Q18
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Electrochemistry question

2025 · 3 Apr · Shift 1 · Q18

JEE MainChemistryElectrochemistryMCQ+4 / −1
Correct order of limiting molar conductivity for cations in water at 298 K is :
  1. A
    H+>Na+>K+>Ca2+>Mg2+\mathrm{H}^{+}\gt \mathrm{Na}^{+}\gt \mathrm{K}^{+}\gt \mathrm{Ca}^{2+}\gt \mathrm{Mg}^{2+}H+>Na+>K+>Ca2+>Mg2+
  2. B
    H+>Ca2+>Mg2+>K+>Na+\mathrm{H}^{+}\gt \mathrm{Ca}^{2+}\gt \mathrm{Mg}^{2+}\gt \mathrm{K}^{+}\gt \mathrm{Na}^{+}H+>Ca2+>Mg2+>K+>Na+
  3. C
    Mg2+>H+>Ca2+>K+>Na+\mathrm{Mg}^{2+}\gt \mathrm{H}^{+}\gt \mathrm{Ca}^{2+}\gt \mathrm{K}^{+}\gt \mathrm{Na}^{+}Mg2+>H+>Ca2+>K+>Na+
  4. D
    H+>Na+>Ca2+>Mg2+>K+\mathrm{H}^{+}\gt \mathrm{Na}^{+}\gt \mathrm{Ca}^{2+}\gt \mathrm{Mg}^{2+}\gt \mathrm{K}^{+}H+>Na+>Ca2+>Mg2+>K+
View written solutionFree

Correct answer: B

  1. Concept used: limiting molar conductivity

    At infinite dilution, the limiting molar conductivity of an ion depends on its ionic mobility in water.

    • H+\mathrm{H^+}H+ has exceptionally high conductivity due to the proton hopping mechanism (Grotthuss mechanism).
    • For other cations, mobility depends on size and extent of hydration.
    • Strongly hydrated ions move more slowly.
  2. Known relative ionic conductivities in water at 298 K298\,\text{K}298K

    The standard limiting ionic molar conductivities are approximately:

    λ∘(H+)≫λ∘(Ca2+)>λ∘(Mg2+)>λ∘(K+)>λ∘(Na+)\lambda^\circ(\mathrm{H^+}) \gg \lambda^\circ(\mathrm{Ca^{2+}}) > \lambda^\circ(\mathrm{Mg^{2+}}) > \lambda^\circ(\mathrm{K^+}) > \lambda^\circ(\mathrm{Na^+})λ∘(H+)≫λ∘(Ca2+)>λ∘(Mg2+)>λ∘(K+)>λ∘(Na+)

    More specifically, typical values are:

    λ∘(H+)≈349.6\lambda^\circ(\mathrm{H^+}) \approx 349.6λ∘(H+)≈349.6 λ∘(Ca2+)≈119\lambda^\circ(\mathrm{Ca^{2+}}) \approx 119λ∘(Ca2+)≈119 λ∘(Mg2+)≈106\lambda^\circ(\mathrm{Mg^{2+}}) \approx 106λ∘(Mg2+)≈106 λ∘(K+)≈73.5\lambda^\circ(\mathrm{K^+}) \approx 73.5λ∘(K+)≈73.5 λ∘(Na+)≈50.1\lambda^\circ(\mathrm{Na^+}) \approx 50.1λ∘(Na+)≈50.1

    Hence,

    H+>Ca2+>Mg2+>K+>Na+\mathrm{H^+} > \mathrm{Ca^{2+}} > \mathrm{Mg^{2+}} > \mathrm{K^+} > \mathrm{Na^+}H+>Ca2+>Mg2+>K+>Na+

  3. Check each option

    • Option A: H+>Na+>K+>Ca2+>Mg2+\mathrm{H^+} > \mathrm{Na^+} > \mathrm{K^+} > \mathrm{Ca^{2+}} > \mathrm{Mg^{2+}}H+>Na+>K+>Ca2+>Mg2+

      This is incorrect because K+>Na+\mathrm{K^+} > \mathrm{Na^+}K+>Na+, and also Ca2+,Mg2+\mathrm{Ca^{2+}}, \mathrm{Mg^{2+}}Ca2+,Mg2+ have higher limiting molar conductivity than K+,Na+\mathrm{K^+}, \mathrm{Na^+}K+,Na+.

    • Option B: H+>Ca2+>Mg2+>K+>Na+\mathrm{H^+} > \mathrm{Ca^{2+}} > \mathrm{Mg^{2+}} > \mathrm{K^+} > \mathrm{Na^+}H+>Ca2+>Mg2+>K+>Na+

      This matches the correct order.

    • Option C: Mg2+>H+>Ca2+>K+>Na+\mathrm{Mg^{2+}} > \mathrm{H^+} > \mathrm{Ca^{2+}} > \mathrm{K^+} > \mathrm{Na^+}Mg2+>H+>Ca2+>K+>Na+

      Incorrect because H+\mathrm{H^+}H+ must be the highest.

    • Option D: H+>Na+>Ca2+>Mg2+>K+\mathrm{H^+} > \mathrm{Na^+} > \mathrm{Ca^{2+}} > \mathrm{Mg^{2+}} > \mathrm{K^+}H+>Na+>Ca2+>Mg2+>K+

      Incorrect because K+>Na+\mathrm{K^+} > \mathrm{Na^+}K+>Na+, and also Ca2+,Mg2+>K+,Na+\mathrm{Ca^{2+}}, \mathrm{Mg^{2+}} > \mathrm{K^+}, \mathrm{Na^+}Ca2+,Mg2+>K+,Na+.

  4. Final answer

    The correct order is:

    H+>Ca2+>Mg2+>K+>Na+\boxed{\mathrm{H^+} > \mathrm{Ca^{2+}} > \mathrm{Mg^{2+}} > \mathrm{K^+} > \mathrm{Na^+}}H+>Ca2+>Mg2+>K+>Na+​

    So the correct option is B.

  5. Comparison with stored correct answer

    Stored correct answer = B

    My derived answer = B

    Therefore, they agree.

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