The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of 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 `5 xx 10^(-7)S`. The pH of the solution is 4. The value of limiting molar conductivity `(Lambda^(@))` of this monobasic acid in aqueous solution is `Z xx 10^(2)S cm^(-1) mol^(-1)`. The value of Z is ......... .

Explanation

Conc `C=0.0015M`, conductance, `G=5xx10^(-7)S`
`l=120cm,a=1cm^(2)`
`wedge_(m)^(c)=(Kxx1000)/("Molarity")=(6xx10^(-5)xx1000)/(0.0015)`
`=40" S "cm^(2)mol^(-1)`
`{:(HA,hArr,H^(+),+,A^(-)),(C,,0,,0),(C-Calpha,,Calpha,,Calpha):}`
`[H^(+)]=Calpha=10^(-4)M" "(becausepH=4)`
`theeforealpha=(10^(-4))/(C)=(10^(-4))/(0.0015)`
`alpha=(wedge_(m)^(c))/(wedge_(m)^(@)) therefore(10^(-4))/(0.0015)=(40)/(wedge_(m)^(@))`
or `wedge_(m)^(@)=(40xx0.0015)/(10^(-4))=6xx10^(2)" S "cm^(2)mol^(-1)`