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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Chemistry
Chapter
Electrochemistry

JIPMER 2015 · Chemistry · Electrochemistry

The conductivity of 0.001028 ~mol ~L^ -1 acetic acid is 4.95 × 10^...

The conductivity of 0.001028 mol L−10.001028 \mathrm{~mol} \mathrm{~L}^{-1} acetic acid is 4.95×10−5 S cm−14.95 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}. Find out its dissociation constant if Λm\Lambda_{\mathrm{m}} for acetic acid is 390.5 S cm−1 mol−1390.5 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}.

Options

  1. A2.18×10−5 mol−1 L−12.18 \times 10^{-5} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}
  2. B1.78×10−5 mol L−11.78 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}
  3. C3.72×10−4 mol L−13.72 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}
  4. D2.37×10−4 mol L−12.37 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}

Correct answer

B. 1.78×10−5 mol L−11.78 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}

Step-by-step solution

Λm=KC=4.95×10−5 S cm−10.001028 mol L−1×1000 cm3 L=48.15 S cm2 mol−1α=ΛmΛm∘=48.15390.5=0.1233K=Cα2(1−α)=0.001028×(0.1233)2(1−0.1233)=1.78×10−5 mol L−1\begin{aligned} \Lambda_{\mathrm{m}}=\frac{K}{C} & =\frac{4.95 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{0.001028 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}} \\ & =48.15 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \alpha & =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{48.15}{390.5}=0.1233 \\ K & =\frac{C \alpha^2}{(1-\alpha)}=\frac{0.001028 \times(0.1233)^2}{(1-0.1233)} \\ & =1.78 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}
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