CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

Question : 35

Total: 35

(a) Conductivity of 2×10−3 M methanoic acid is 8 ×10−5 S cm−1. Calculate its molar conductivity and degree of dissociation if ∧m0 for methanoic acid is 404 S cm2 mol−1.
(b) Calculate the ΔG0 and log‌Kc for the given reaction at 298 K :Ni(s)+2Ag(aq)+⇌Ni(laq)2++2Ag(s)
Given : E0Ni‌2+∕Ni=−0.25V, E0Ag+∕Ag=+0.80V
1F=96500Cmol−1.

Solution: 👈: Video Solution

(a) Molar conductivity
Λm= κ×1000C =

8×10−5 S cm−1×1000

2×10−3 mol L−1

8×10−2

2×10−3

=40 S cm2 mol−1
Degree of dissociation

Λm

Λ°m

404

=0.099
(b) Ni────▸Ni2+(E0=−0.25V) ( Oxidation half)
2Ag+────▸2Ag(E0=0.80V) (Reduction half)
E°=Ec−Ea =0.80−(−0.25)=1.05V
ΔG=nFE°
=2×96500×10.5
=202.650 J mol−1
=202.650 kJ mol−1
E°cell =

0.0591

‌log‌ ‌Kc
log‌Kc=

1.05×2

0.0591

=35.53
By taking Antilog
Antilog 35.35
=1053×3.38
∵So Kc=3.38×1053

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