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Q.

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 electrode is 120 cm with an area of cross section of 1 cm2 . The conductance of this solution was found to be5×10−7S . The pH of the solution is 4. The value of limiting molar conductivity Λm0 of this weak monobasic acid in aqueous solution is Z×102S cm−1 mol−1. The value of Z is

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answer is 6.

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Detailed Solution

K=G× cell constant k=5×10−7S×1201cmcm2=600×10−7Scm−1Λm=600×10−7×1000C=600×10−7×10000.0015M=40pH=4(HT)=Cα=10−4α=115ΛMΛm0=αΛm0=Λmα=401/15=600600S−cm2mol−12×102=6×102
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