Resistance of a conductivity cell filled with 0.1

Asked by admin @ 29/08/2021 in Chemistry viewed by 277 People

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Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 ohm. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 ohm, calculate the conductivityand molar conductivity of 0.02 M KCl solution. The conductivity of 0.1 M KCl solution is 1.29 S/m.

Answered by admin @ 29/08/2021

Answer :

$\rm{Conductivity=0.248\:S\:m^{-1}}$
$\rm{Molar\:conductivity=1.24\times{}10^{-2}\:S\:m^2\:mol^{-1}}$

Step-by-step explanation:

Given that, for 0.1 M KCl solution,

R = 100 Ω.
κ = 1.29 S/m.

Also, for 0.02 M KCl solution,

R = 520 Ω.
κ = ?

$\hrulefill$

We know, Cell constant = Conductiviy × Resistance.

∴ For 0.1 M KCl solution,

$\rm{Cell\:constant=1.29\:S\:m^{-1}\times{}100\:\Omega}$

$\rm{Cell\:constant=129\:m^{-1}=1.29\:cm^{-1}}$

$\rule{350}{3}$

$\rm{Now,\:Conductivity\:of\:0.02\:M\:KCl\:solution=\dfrac{Cell\:constant}{Resistance}}$

$\rm{\kappa=\dfrac{129\:m^{-1}}{520\:\Omega}=0.248\:\Omega^{-1}\:m^{-1}=}\bold{0.248\:S\:m^{-1}}$

Again, concentration of the solution = 0.02 M = $\rm{20\:mol\:m^{-3}}$

$\rule{350}{3}$

$\rm{Molar\:conductivity=\dfrac{\kappa}{c_m}=\dfrac{0.248\:S\:m^{-1}}{20\:mol\:m^{-3}}}$

$\rm{Molar\:conductivity=0.0124\:S\:m^2\:mol^{-1}=}\bold{1.24\times{}10^{-2}\:S\:m^2\:mol^{-1}}$

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