Question

The measured resistance of a conductivity cell containing `7.5xx10^(-3)` M solution of KCl at `43^(@)` C was 1005 ohms. Calculate (a) Specific conductnce (b) Molar conductance of the solution. Given cell constant =1.25 `cm^(-1)`.

Answer 1

Resistance (R )=1005 `Lambda`
Conductance (C )`=(1)/(R )=(1)/(1005 " ohm")`
Specific conductance (k) `="Conductance"xx"Cell constant"`.
`=(1)/(1005 " ohm")xx1.25 cm^(-1)=1.244xx10^(-3) " ohm"^(-1)cm^(-1)`
Step II. Calculation of the molar conductance
Molar concentration (C ) ` =7.5xx10^(-3)" mol " L^(-1)`
`=(7.5xx10^(-3) " mol")/(1L)=(7.5xx10^(-3)" mol")/(10^(3) cm^(3))=7.5xx10^(-6) " mol " cm^(-3)`
Molar conductance `(Lambda_(m))=(k)/(C )=((1.244xx10^(-3)" ohm "^(-1) cm^(-1))/(7.5xx10^(-6) "mol" cm^(-3)))=165.87 " ohm"^(-1) cm^(2) "mol"^(-1)`