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

6

Explanation

`[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`