0.30 Conductivity of 2.5x10-M Methanoic acid is 5.25 x 105 SCm. Calculate its molar conductivity
and degree of dissociation.
Given 2 (H') = 349.5 SCmº mol'' and a(HC00")= 50.5 Scm moll
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ANSWER
We know molar conductivity, (λ
m
)=
concentration(c)
1000×conductivity(k)
∴λ
m
=
2.5×10
−4
1000×5.25×10
−5
=210Scm
2
mol
−1
λ
CHCOOH
0
=λ
H
+
0
+λ
(HCOO
−
)
0
=(349.5+50.5)=400Scm
2
mol
−1
∴α=
λ
0
λ
m
=
400
210
=0.52
or, α=52.5 %
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