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Resistance of 0.05 M electrolytic solution at 298K temperature is 30.0 Omega. The cross sectional area of conductivity cell having Pt electrode is 3.8cm^(2) and distance between two electrode is 1.5 cm, then what is the molar conductivity of electrolytic solution ? |
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Answer» SOLUTION :Conductivity constant for cell : `G^(**)=(l)/(A)=(1.5cm)/(3.8cm^(2))=0.3947cm^(-1)` Where, R=resistance of solution=`30.0Omega` c=concentration of solution=0.05 mol `L^(-1)` Conductivity of solution : `kappa=(G^(**))/(R)=(1.5cm)/(3.8cm^(2))xx(1)/(30Omega)` `=0.013158cm^(-1)Omega^(-1)~~0.0132" S "CM^(-1)` Molar conductivity : `Lamda_(m)=(1000kappa)/(c)=(1000xx0.0132" S "cm^(-1))/(0.05" mol "cm^(-3))` `=264" S cm"^(2)mol^(-1)` |
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