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

Answer

`rho=87.135"ohm cm or "87.135xx10^(-2) "ohm m"; k=0.01148 "S cm"^(-1) or 1.148 Sm^(-1): wedge_(m)=229.6 "S cm"^(2) "mol"^(-1) or 229.6 xx 10^(-4) m^(2)"mol"^(-1)`

Explanation

Radius `r=("diameter")/(2) =(1 cm)/(2) =0.5 cm, "length", l=50 cm`
or `0.5m, A = pi r^(2) =3.142 xx (0.5 cm)^(2)=0.785 cm^(2)`
or `0.785xx10^(-4)m^(2)`
Resistivity, `rho=(R.A)/(l)=(5.55xx10^(3)"ohm" xx 0.785 cm^(2))/(50cm)`
= 87.135 ohm cm or 87.135 `xx 10^(-2)` ohm m
Conductivity, `k=(1)/(rho)=(1)/(87.135" ohm cm")`
`=0.01148" ohm"^(-1)cm^(-1) or 1.485"S m"^(-1)`
`wedge_(m)=(k xx 1000 cm^(3))/("Molarity")=(0.01148xx1000)/(0.05)`
`=229.65" S cm"^(2)"mol"^(-1) or 229.6 xx 10^(-4)"S m"^(2)"mol"^(-1)`