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

`C=0.0015M" "l=120cm`
`G=5xx10^(-7)s" "a=1cm^(2)`
`G=Kxx(a)/(l)`
`5xx10^(-7)=Kxx(1)/(120)`
`K=6xx10^(-5)S" "cm^-1`
`(^^)_(m)^(c)=(Kxx1000)/(M)=(6xx10^(-5)xx1000)/(0.0015)`
`pH=4implies[H^(+)]=10^(-4)=calpha=0.0015alpha`
`alpha=(10^(-4))/(0.0015)" "alpha=((^^)_(m)^(c))/((^^)_(m)^(0))impleis(10^(-4))/(0.0015)=((6xx10^(-5)xx1000)/(0.0015))/((^^)_(m)^(0))`
`(^^)_(m)^(0)=6xx10^(2)s" "cm^(2)"mole"-1`