Question

a b) when a conductivity cell was filled with o. 1 mol - KCl, it has resistence of loor. If the same cell was filled with an aqueous solution of 0.02 mol L of KCl, resistence become 520r Hence Atlet find out the conductivity and molar conductivity mol Altx Xem L Kon[ the conductivity of 0.1 mol L-1 kai solution is 1.29 s/m]. of 0.02​

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