The elctrical resistance of a column of `0.05 M N aOH ` solution of diameter `1 cm` and length `50 cm` is `5.55 xx 10^(3)ohm`. Calculate its resisteivity , conductivity, and molar conductivity.

Options

A. `87.135 Omega,0.01148 S cm^(-1),229.6 S cm^(2)" mol"^(-1)`
B. `77.123 Omega,0.241 S cm^(-1),119.5 S cm^(2)" mol"^(-1)`
C. `92.29 Omega,0.124 S cm^(-1),272.3 S cm^(2)" mol"^(-1)`
D. `83.92 Omega,0.1192 S cm^(-1),220.3 S cm^(2)" mol"^(-1)`

Answer

A

Explanation

`A =pir^(2)=3.14xx0.5^(2)cm^(2)`
`=0.785cm^(2)=0.785xx10^(-4)cm^(2)`
`l=50cm=0.5m`
`R=(rhol)/(A)or rho=(RA)/(ll)=`
`(5.55xx10^(3)Omegaxx0.785cm^(2))/(50cm)=87.135Omegacm`
`"Conductivity"=K=(l)/(rho)=((1)/(87.135))"S cm"^(-1)`
`=(0.01148"S cm"^(-1)xx1000cm^(3)L^(-1))/(0.05"molL"^(-1))=229.6"Scm"^(2)"mol"^(-1)`
If want to calculate the values of different quantities in terms of m instead of cm .
`rho=(RT)/(l)=(5.55xx10^(3)Omegaxx0.785xx10^(-4)m^(2))/(0.5m)=87.135xx10^(-2)Omegam`
`K=(1)/(rho)=(100)/(87.135)Omegam=1.148"S m"^(-1)`
and
`Lambda_(m)=(K)/(C)=(1.148Sm^(-1))/(50"mom m"^(-3))=229.6xx10^(-4)"Sm"^(2)mol^(-1)`