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Chemistry · Electrochemistry

JEE Advanced 2023 — Paper 1 — Question 35

Plotting 1/Λm1 / \Lambda_{\mathrm{m}} against cΛm\mathrm{c} \Lambda_{\mathrm{m}} for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y -axis intercept of P and slope of S . The ratio P/S\mathrm{P} / \mathrm{S} is [ Λm=\Lambda_{\mathrm{m}}= molar conductivity Λm0=\Lambda_{\mathrm{m}}^{0}= limiting molar conductivity c=\mathrm{c}= molar concentration Ka=\mathrm{K}_{\mathrm{a}}= dissociation constant of HX]

  1. Option A:

    KaΛm0\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{0}

    Correct
  2. Option B:

    KaΛm0/2\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{0} / 2

  3. Option C:

    2 KaΛm02 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{0}

  4. Option D:

    1/(KaΛm0)1 /\left(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{0}\right)

Answer: A

Step-by-step solution

Ka=C(λmλm0)2(1−λmλm0)\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C}\left(\frac{\lambda_{\mathrm{m}}}{\lambda_{\mathrm{m}}^{0}}\right)^{2}}{\left(1-\frac{\lambda_{\mathrm{m}}}{\lambda_{\mathrm{m}}^{0}}\right)} Ka=Cλm2λm0λm0−λm\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \frac{\lambda_{\mathrm{m}}^{2}}{\lambda_{\mathrm{m}}^{0}}}{\lambda_{\mathrm{m}}^{0}-\lambda_{\mathrm{m}}} Ka=Cλm2λm0(λm0−λm)\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \lambda_{\mathrm{m}}^{2}}{\lambda_{\mathrm{m}}^{0}\left(\lambda_{\mathrm{m}}^{0}-\lambda_{\mathrm{m}}\right)} Ka(λm0)2−Kaλmλm0=Cλm2\mathrm{K}_{\mathrm{a}}\left(\lambda_{\mathrm{m}}^{0}\right)^{2}-\mathrm{K}_{\mathrm{a}} \lambda_{\mathrm{m}} \lambda_{\mathrm{m}}^{0}=\mathrm{C} \lambda_{\mathrm{m}}^{2} Ka(λm0)2λm−Kaλm0=Cλm\mathrm{K}_{\mathrm{a}} \frac{\left(\lambda_{\mathrm{m}}^{0}\right)^{2}}{\lambda_{\mathrm{m}}}-\mathrm{K}_{\mathrm{a}} \lambda_{\mathrm{m}}^{0}=\mathrm{C} \lambda_{\mathrm{m}} 1λm=+(cλmKa(λM0)2)+kaλm0 Ka(λm0)2\frac{1}{\lambda_{\mathrm{m}}}=+\left(\frac{\mathrm{c} \lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\lambda_{\mathrm{M}}^{0}\right)^{2}}\right)+\frac{\mathrm{k}_{\mathrm{a}} \lambda_{\mathrm{m}}^{0}}{\mathrm{~K}_{\mathrm{a}}\left(\lambda_{\mathrm{m}}^{0}\right)^{2}} 1λm=cλmKa(λm0)2+1λm0\frac{1}{\lambda_{\mathrm{m}}}=\frac{\mathrm{c} \lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\lambda_{\mathrm{m}}^{0}\right)^{2}}+\frac{1}{\lambda_{\mathrm{m}}^{0}} S=1 Ka(λm0)2;P=1λm0\mathrm{S}=\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\lambda_{\mathrm{m}}^{0}\right)^{2}} ; \mathrm{P}=\frac{1}{\lambda_{\mathrm{m}}^{0}} PS=Kaλm0\frac{\mathrm{P}}{\mathrm{S}}=\mathrm{K}_{\mathrm{a}} \lambda_{\mathrm{m}}^{0}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law