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AIIMS 2016
Exam
AIIMS
Year
2016
Subject
Chemistry
Chapter
Electrochemistry

AIIMS 2016 · Chemistry · Electrochemistry

A conductivity cell has a cell constant of 0.5 ~cm^ -1 . This cell...

A conductivity cell has a cell constant of 0.5 cm−10.5 \mathrm{~cm}^{-1}. This cell when filled with 0.01M0.01 \mathrm{M} NaCl\mathrm{NaCl} solution has a resistance of 384ohms384 \mathrm{ohms} at 25∘C25^{\circ} \mathrm{C}. Calculate the equivalent conductance of the given solution.

Options

  1. A130.2Ω−1 cm2( geq)−1130.2 \Omega^{-1} \mathrm{~cm}^2(\mathrm{~g} \mathrm{eq})^{-1}
  2. B137.4Ω−1 cm2( geq)−1137.4 \Omega^{-1} \mathrm{~cm}^2(\mathrm{~g} \mathrm{eq})^{-1}
  3. C154.6Ω−1 cm2( geq)−1154.6 \Omega^{-1} \mathrm{~cm}^2(\mathrm{~g} \mathrm{eq})^{-1}
  4. D169.2Ω−1 cm2( geq)−1169.2 \Omega^{-1} \mathrm{~cm}^2(\mathrm{~g} \mathrm{eq})^{-1}

Correct answer

A. 130.2Ω−1 cm2( geq)−1130.2 \Omega^{-1} \mathrm{~cm}^2(\mathrm{~g} \mathrm{eq})^{-1}

Step-by-step solution

 Equivalent conductance, Λeq=κ×1000 Normality  where, κ (Specific conductance) =C×la=1R×la=1384×0.5=1.302×10−3ohm−1cm−1∴Λeq=1.302×10−3×10000.01=130.2ohm−1cm2(geq)−1\begin{aligned} & \text{ Equivalent conductance, } \Lambda_{e q}=\frac{\kappa \times 1000}{\text{ Normality }} \\ & \text{ where, } \kappa \text{ (Specific conductance) }=C \times \frac{l}{a} \\ & \quad=\frac{1}{R} \times \frac{l}{a}=\frac{1}{384} \times 0.5 \\ & \quad=1.302 \times 10^{-3} \mathrm{ohm}^{-1} \mathrm{ cm}^{-1} \\ & \therefore \quad \Lambda_{e q}=\frac{1.302 \times 10^{-3} \times 1000}{0.01} \\ & \quad=130.2 \mathrm{ohm}^{-1} \mathrm{ cm}^2(\mathrm{ g} \mathrm{eq})^{-1}\end{aligned}
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