The molar conductivities ΛNaOAco and ΛHClo at infinite dilution in water at 25oC are 91.0 and 426.2 S cm2/mol respectively. To calculate ΛCH3COOHo, the additional value required is:
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a
ΛH2Oo
b
ΛKClo
c
ΛNaOHo
d
ΛNaClo
answer is D.
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Detailed Solution
According to Kohlrausch's law; ΛCH3COOHo=ΛCH3COONao+ΛHClo−ΛNaClo