Question

the molar conductivity of 0.001 M acetic acid is 4.95 ×10^-5 S cm^2 mol^ -1 .calculate the degree of dissociation at this concentration if the limiting molar conductivity for H+ is 340×10^-5 S cm^2 mol^-1 and for CH3COO- is 50.5×10^-5 S cm^2 mol^-1
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Answer 1

Explanation:

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