The equivalent conductivities of sodium benzoate and HCl at 250C are 82.4 and 426.16 Scm2equiv–1 respectively. If the transport numbers of benzoate and H+ ions are 0.392 and 0.821 in the above mentioned compounds, find the equivalent conductivity of benzoic acid at 250C.
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Answer:
CH
3
COONa+HCl⟶CH
3
COOH+NaCl
λ
eq(CH
3
COOH)
∞
=λ
eq(CH
3
COONa)
∞
+λ
eq(HCl)
∞
−λ
eq(NaCl)
∞
λ
eq(CH
3
COOH)
∞
=91+425.16−126.8=389.71 ohm
−1
cm
2
eq
where λ
eq
∞
= Equivalence conductivity at infinitie dilution.
Hence, the correct answer is (C)
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