Question

Resistance of a conductivity cell filled with 0.1 mol L KCl solution is Example
100 22. If the resistance of the same cell when filled with 0.02 mol L-
KCl solution is 5202, calculate the conductivity and molar conductivity
of 0.02 mol L KCl solution. The conductivity of 0.1 mol L-KCI
solution is 1.29 S/m.​

Answer 1

Answer:

For 0.1molL

−1

KCl solution,

Conductivity , k=1.29×10

−2

Ω

−1

cm

−1

, Resistance , R=100W

Cell constant = Conductivity × resistance

=1.29×10

−2

Ω

−1

cm

−1

×100Ω=1.29cm

−1

For 0.02molL

−1

solution,

Resistance =520Ω, Cell constant =1.29cm

−1

,

Conductivity , k=

Resistance

Cell constant

=

520Ω

129

=0.00248Ω

−1

cm

−1

Molar conductivity , Λ

m

=

Molarity

Conductivity(k)×1000cm

3

L

−1

=

0.02molL

−1

0.00248Ω

−1

×1000cm

3

L

−1

=124Ω

−1

cm

2

mol

−1

Explanation:

100% correct answer

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