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The electrical resistance of a column 0....

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

Text Solution

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`A =pir^(2) =31.4 xx0.5^(5) cm^(2) =0.758 cm^(2) =0.758 xx10^(-4)m^(2)`
`l=50 cm =0.5m`
`R= (rhol)/(A)or rho =(RA)/(l )=(5.55 xx10^(3) Omegaxx0.785cm^(2))/(50cm)=87.135 Omega cm`
Conductivity `R= (rhol)/(A)or rho =(RA)/(l )=(5.55 xx10^(3) Omegaxx0.785cm^(2))/(50cm)=87.135 Omega cm`
Molar conductivity, `=k =1/rho=((1)/(87.135))S cm^(-1)=0.01148 S cm^(-1)`
`^^_(m) =(kxx 1000)/( c) cm^(3) L^(-1)`
`=(0.01148 S cm^(-1)xx1000cm^(3) L ^(-1))/(0.05 mol L^(-1))=229.6 S cm ^(2) mol ^(-1)`
If we want to calculate the calues of different quantities in term of 'm' instead of 'cm'.
`rho =(RA)/(l) =(5.35 xx10^(3)Omega xx0.785 xx10^(-4)m^(2))/(0.5m)`
`=87.135 xx10^(-2) Omegam`
`k=1/rho=(100)/( 87.135) Omegam =1.148 S m^(-1)`
`and ^^_(m) =k/c =(1.148 Sm^(-1))/(50 mol m^(-3))=299.6xx10^(-4) Sm^(2)mol ^(-1)`
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