A 0.05 M NaOH solution offered a resistance of 31.6 `Omega` in a conductivity cell at 298 K. If the cell constant of the conductivity cell is 0.367 `cm^(-1)`, find out the specific and molar conductance of the sodium hydroxide solution.

Explanation

Step I. Calculate of the specific conductance
Resistance (R ) =31.6 `Omega`
`:."Conductance" (C )=(1)/(R )=(1)/(31.6 " ohm")=0.0316 " ohm"^(-1)`
`"Specific conductance" (k)= "Conductance" xx"cell constant" `
`=(0.0316 " ohm"^(-1))xx(0.367 cm^(-1))`
`=0.0116 " ohm"^(-1) cm^(-1)`.
Step II. Calculate of the conductance.
Molar concentration, (C ) =0.05 M =0.05 mol `L^(-1)`
`=(0.05 " mol")/(1L)=(0.05" mol")/(10^(3) cm^(3))=0.05xx10^(-3) " mol " cm^(-3)`
`"Molar conductance" (Lambda_(m))=(k)/(C )=((0.0116 " oh "m^(-1)cm^(-1)))/((0.05xx10^(-3) " mol " cm^(-3)))=232 " oh "m^(-1) cm^(-1)cm^(2) "mol"^(-1)`.