Question

Molar conductivity of 0.01 molar acetic acid at 25 degree Celsius is 16.5 ohm inverse centimetre Square mol inverse at 0.01 concentration calculate its degree of dissociation in 0.01 mole and associate at three 90.7 ohm inverse Square mall inverse​

Answer 1

Explanation:

Data:-

Concentration(C)=0.01moldm

−3

=10molm

−3

.

Conductivity(K)=19.5×10

−5

Scm

−1

=19.5×10

−3

Sm

−1

Limiting molar conductivity(∧

∘

m)=390Scm

2

mol

−1

=390×10

−4

Sm

2

mol

−1

Solution:-

(i) Molar Conductivity∧

m

=

C

K

=

10molm

−3

19.5×10

−3

Sm

−1

∧

m

=19.5×10

−4

Sm

2

mol

−1

.

(ii) Degree of dissociation(α)=

∧

m

∘

∧

m

=

390×10

−4

Sm

2

mol

−1

19.5×10

−4

Sm

2

mol

−1

∴α=0.05