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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Chemistry
Chapter
Electrochemistry

JIPMER 2018 · Chemistry · Electrochemistry

If molar conductivity of Ca^ 2+ and Cl^ - ions are 119 and 71 Sm^2...

If molar conductivity of Ca2+\mathrm{Ca}^{2+} and Cl−\mathrm{Cl}^{-}ions are 119 and 71Sm2 mol−171 \mathrm{Sm}^2 \mathrm{~mol}^{-1} respectively, then the molar conductivity of CaCl2\mathrm{CaCl}_2 at infinite dilution is

Options

  1. A215 S cm2 mol−1215 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
  2. B340 S cm2 mol−1340 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
  3. C126 S cm2 mol−1126 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
  4. D261 S cm2 mol−1261 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Correct answer

D. 261 S cm2 mol−1261 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Step-by-step solution

Given, Λ∘Ca2+=119 S cm2 mol−1Λ∘Cl−=71 S cm2 mol−1\begin{array}{r}\Lambda^{\circ} \mathrm{Ca}^{2+}=119 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \Lambda^{\circ} \mathrm{Cl}^{-}=71 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{array} As we know that, Λ∘CaCl2=ΛCa2+∘+2×Λ∘Cl−∴Λ∘CaCl2=119+2×71Λ∘CaCl2=261 S cm2 mol−1\begin{array}{ll} & \Lambda^{\circ} \mathrm{CaCl}_2=\Lambda_{\mathrm{Ca}^2+}^{\circ}+2 \times \Lambda^{\circ}{ }_{\mathrm{Cl}^{-}} \\ \therefore \quad & \Lambda^{\circ} \mathrm{CaCl}_2=119+2 \times 71 \\ & \Lambda^{\circ} \mathrm{CaCl}_2=261 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{array}
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