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Chemistry · Electrochemistry

JEE Main 2024 — 4 April, Shift 2 — Question 72

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

  1. Option A:

    Scm2 mol−3/2 L1/2\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{1 / 2}

    Correct
  2. Option B:

    Scm2 mol−1 L1/2\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \mathrm{~L}^{1 / 2}

  3. Option C:

    Scm2 mol−3/2 L\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-3 / 2} \mathrm{~L}

  4. Option D:

    Scm2 mol−3/2 L−1/2\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{-1 / 2}

Answer: A

Step-by-step solution

Λm=Λm∘−AC\quad \Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{C}}

Units of AC=Scm2\mathrm{A} \sqrt{\mathrm{C}}=\mathrm{S} \mathrm{cm}^{2} mole −1^{-1}

Uits of A=Scm2\mathrm{A}=\mathrm{S} \mathrm{cm}^{2} mole −3/2 L1/2^{-3 / 2} \mathrm{~L}^{1 / 2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law

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