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Electrochemistry question

2023 · Shift 1 · Q5
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  5. /2023 · Shift 1 · Q5

Electrochemistry question

2023 · Shift 1 · Q5

JEE AdvancedChemistryElectrochemistryMCQ+3 / −1
Plotting 1/Λm1 / \Lambda_{\mathrm{m}}1/Λm​ against cΛm\mathrm{c} \Lambda_{\mathrm{m}}cΛm​ for aqueous solutions of a monobasic weak acid (HX)(\mathrm{HX})(HX) resulted in a straight line with y\mathrm{y}y-axis intercept of P\mathrm{P}P and slope of S\mathrm{S}S. The ratio P/S\mathrm{P} / \mathrm{S}P/S is [Λm= molar conductivity Λmo= limiting molar conductivity c= molar concentration Ka= dissociation constant of HX]\begin{aligned} & {\left[\Lambda_{\mathrm{m}}=\right.\text { molar conductivity }} \\\\ & \Lambda_{\mathrm{m}}^{\mathrm{o}}=\text { limiting molar conductivity } \\\\ & \mathrm{c}=\text { molar concentration } \\\\ & \left.\mathrm{K}_{\mathrm{a}}=\text { dissociation constant of } \mathrm{HX}\right] \end{aligned}​[Λm​= molar conductivity Λmo​= limiting molar conductivity c= molar concentration Ka​= dissociation constant of HX]​
  1. A
    KaΛmo\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}Ka​Λmo​
  2. B
    KaΛmo/2\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}} / 2Ka​Λmo​/2
  3. C
    2 KaΛmo2 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}2 Ka​Λmo​
  4. D
    1/(KaΛmo)1 /\left(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}\right)1/(Ka​Λmo​)
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