Question

The electrical resistance of a column of `0.05 " mol " L^(-1)` NaOH solution of diameter 1 cm and length 50 cm is `5.55xx10^(3)" ohm "`.Calculate its resistivity,conductivity and molar conductivity.

Answer 1

Step I : Calculate of resistivity or specific resistance
Resistance (R )`=5.55xx10^(3) Lambda`
Area of cross-section of column (a)`=pir^(2)=3.14xx(0.5)^(2) cm^(2)=0.785 cm^(2)`
Length of the column(l)=50 cm
Resistivity (p)`=(Rxxa)/(l)=((5.55xx10^(3)" oh"m)xx(0.785 cm^(2)))/((50 cm))`
`=87.135 "ohm" cm`.
Step II : Calculation of conductivity or specific conductance
Specific conductance (k)`=(1)/(rho)=(1)/((87.135 " ohm "cm))`
`= 0.01148 "oh"m^(-1) cm^(-1)`.
Step III : Calculation of molar conductivity
Molar concentration (C ) `=0.05 " mol "L^(-1)=(5xx10^(-2)" mol")/(1L)=((5xx1 0^(-2)mol))/((10^(3) cm^(3))`
`=5xx10^(-5) " mol" cm^(-3)`
Molar conductivity `(Lambda_(m))=(k)/(C )=((0.01148 " oh"m^(-1)cm^(-1)))/((5xx10^(-5) " mol " cm^(-3)))`
`229.6 "ohm"^(-1)cm^(2) mol^(-1)`.