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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020
ChemistryEasynumerical

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt- electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1cm2. The conductance of this solution was found to be5×10-7S. The pH of the solution is 4. The value of limiting molar conductivityΛmoif this weak monobasic acid in aqueous solution isZ×102Scm2mol-1. The value of Z is

Answer:
6.00
Solution:

[ H + ] =  10 −pH  =  10 −4  M
ΛM=k×1000M
=G×la×1000M
=5×10-7×1201×10000.0015
=40  Scm 2 mol −1
Now,H+=Cα=0.0015×ΛmΛm∞
Λ m ∞  = Z ×  10 2  =  0.0015 × 40 10 −4
On solving Z = 6.

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed
⏱ 2mℹ️ Source: PYQ_2020

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