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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 3 April · Morning Shift · Chemistry · Electrochemistry

Correct order of limiting molar conductivity for cations in water...

Correct order of limiting molar conductivity for cations in water at 298 K is :

Options

  1. AH+>Na+>K+>Ca2+>Mg2+\mathrm{H}^{+} \gt \mathrm{Na}^{+} \gt \mathrm{K}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+}
  2. BH+>Ca2+>Mg2+>K+>Na+\mathrm{H}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}
  3. CMg2+>H+>Ca2+>K+>Na+\mathrm{Mg}^{2+} \gt \mathrm{H}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}
  4. DH+>Na+>Ca2+>Mg2+>K+\mathrm{H}^{+} \gt \mathrm{Na}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+} \gt \mathrm{K}^{+}

Correct answer

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

Step-by-step solution

Limiting Molar Conductivities of Ions:
- H⊕:349.8Scm2 mol−1\stackrel{\oplus}{\mathrm{H}}: 349.8 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
- Na+:50.11Scm2 mol−1\mathrm{Na}^{+}: 50.11 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
- K+:73.52Scm2 mol−1\mathrm{K}^{+}: 73.52 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
- Ca+2:119Scm2 mol−1\mathrm{Ca}^{+2}: 119 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
- Mg+2:106.12Scm2 mol−1\mathrm{Mg}^{+2}: 106.12 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
Therefore correct order of limiting molar conductivity of cations will be -
H⊕>Ca+2>Mg+2>K+>Na+\stackrel{\oplus}{\mathrm{H}} \gt \mathrm{Ca}^{+2} \gt \mathrm{Mg}^{+2} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}
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