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Question: Each of the two platinum electrodes having the area \(64m{{m}^{2}}\) of a conductivity cell are sepa...

Each of the two platinum electrodes having the area 64mm264m{{m}^{2}} of a conductivity cell are separated by 8mm.The resistance of the cell containing 7.5×10−37.5\times {{10}^{-3}}M KCl solution at 298K is 1250 ohm. Calculate:
(i)Cell constant
(ii)Conductivity and
(ii) Molar conductivity

Explanation

Solution

Cell constant, conductivity and molar conductivity are the terms of an electro-chemical cell all these are inter-related to each other. Calculating cell constant can give conductivity of a given cell knowing which is used to determine molar conductivity by one mole of the electrolyte.

Complete answer:
(i) Cell constant is defined as the ratio of distance between the two parallel electrodes in a conductivity cell (l) to the area of cross-section of the electrode (A). The formula for cell constant is l/A.
Given the distance between the electrodes (l) is 8mm and area of cross-section of electrode is 64mm264m{{m}^{2}}sq.mm, upon substituting the values in the above formula we get,
Cell constant =lA=864=0.125mm−1\dfrac{l}{A}=\dfrac{8}{64}=0.125m{{m}^{-1}}
(ii) Conductivity is defined as the reciprocal of specific resistance or resistivity of a given cell. Resistivity is proportionality constant for the resistance offered by a cell. Conductivity is denoted by the symbol κ\kappa(kappa) and the specific resistance by ρ\rho(rho). The formula of conductivity is,
κ=1ρ\kappa =\dfrac{1}{\rho }
To calculate conductivity let us first find out resistivity.
The formula for resistivity is,
ρ=RAl\rho =R\dfrac{A}{l}
The given value of ‘R’ resistance is 1250 ohm and we know that l= 8 mm, A=64 sq.mm,lA=864=0.125mm−1\dfrac{l}{A}=\dfrac{8}{64}=0.125m{{m}^{-1}}
Upon substituting these value in the above equation,
ρ=1250×648\rho =1250\times \dfrac{64}{8}
ρ=104ohm mm\rho ={{10}^{4}}ohm\,mm
Substituting these values in the equation of conductivity and we get,
κ=1104=10−4Sm−1\kappa =\dfrac{1}{{{10}^{4}}}={{10}^{-4}}S{{m}^{-1}}
(iii) Molar conductivity is the conductivity offered by all the ions obtained from one mole of an electrolyte in a solution of given concentration.
It is denoted by the symbolλm{{\lambda }_{m}}, and the equation is,
λm=1000κM{{\lambda }_{m}}=\dfrac{1000\kappa }{M}
The given concentration of is 7.5×10−37.5\times {{10}^{-3}}
The conductivity is10−4Sm−1{{10}^{-4}}S{{m}^{-1}}
Let us substitute the values in the above formula, we get,
λm=1000×10−47.5×10−3{{\lambda }_{m}}=\dfrac{1000\times {{10}^{-4}}}{7.5\times {{10}^{-3}}}
λm=1.333×103Sm−1mol−1{{\lambda }_{m}}=1.333\times {{10}^{3}}S{{m}^{-1}}mo{{l}^{-1}}

Note:
The given data in the question is resistance offered by the cell which is used to calculate specific resistance. Specific resistance is the reciprocal of conductivity and conductivity is in turn directly related to molar conductivity. So resistance itself is enough to calculate molar conductivity provided the cell constant and concentration are known.