Q12 / 90 · JEE Main 2024 (04 Apr Shift 2) · Chemistry

For a strong electrolyte, a plot of molar conductivity against (concentration) \({ }^{1 / 2}\) is a straight line, with a negative slope, the correct unit for the slope is

  • A
    \(\mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{-1 / 2}\)
  • Correct answer
    B
    \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{1 / 2}\)
  • C
    \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}\)
  • D
    \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1} \mathrm{~L}^{1 / 2}\)

Solution

$\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{C}}$
Units of $\mathrm{A} \sqrt{\mathrm{C}}=\mathrm{S} \mathrm{cm}^2 \mathrm{~mole}^{-1}$ Uits of $\mathrm{A}=\mathrm{S} \mathrm{cm}^2 \mathrm{~mole}^{-3 / 2} \mathrm{~L}^{1 / 2}$

Hence, the correct answer is option: 'B'.

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