Correct order of limiting molar conductivity for cations in water at 298 K is :
- $\mathrm{H}^{+} \gt \mathrm{Na}^{+} \gt \mathrm{K}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+}$
- $\mathrm{H}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}$
- $\mathrm{Mg}^{2+} \gt \mathrm{H}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}$
- $\mathrm{H}^{+} \gt \mathrm{Na}^{+} \gt \mathrm{Ca}^{2+} \gt \mathrm{Mg}^{2+} \gt \mathrm{K}^{+}$
Solution
- $\stackrel{\oplus}{\mathrm{H}}: 349.8 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
- $\mathrm{Na}^{+}: 50.11 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
- $\mathrm{K}^{+}: 73.52 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
- $\mathrm{Ca}^{+2}: 119 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
- $\mathrm{Mg}^{+2}: 106.12 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
Therefore correct order of limiting molar conductivity of cations will be -
$\stackrel{\oplus}{\mathrm{H}} \gt \mathrm{Ca}^{+2} \gt \mathrm{Mg}^{+2} \gt \mathrm{K}^{+} \gt \mathrm{Na}^{+}$
Asked in: JEE Main 2025 (03 Apr Shift 1)