The conductance of 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 `1 cm^(2)`. The conductance of this solution was found to be `5xx10^(-7) S`. The pH of the solution is 4. Calculate the value of limiting molar conductivity.

The conductance of 0.0015 M aqueous solution of a weak monobasic acid

1.	The conductance of 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 `1 cm^(2)`. The conductance of this solution was found to be `5xx10^(-7) S`. The pH of the solution is 4. Calculate the value of limiting molar conductivity.
Answer» Correct Answer - 6 `[H^(+)]=10^(-pH)=10^(-4)M` `Lambda_(m)=kappaxx1000/M` `=[C xx l/A]xx1000/M` `=5xx10^(-7)xx120/1xx1000/0.0015=40" S cm"^(2)"mol"^(-1)` `[H^(+)]=c alpha=0.0015 xx Lambda_(m)/Lambda_(m)^(@)` `10^(-4)=0.0015xx40/(Zxx 10^(2))` `Z=6`

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