NEETChemistryElectrochemistryMCQ+4 / −1
Molar conductivities at infinite dilution of NaCl, Hcl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol1 respectively. for CH3COOH will be
- A425.5 S cm2 mol1
- B180.5 S cm2 mol1
- C290.8 S cm2 mol1
- D390.5 S cm2 mol1
View written solutionFree
Correct answer: D
oCH3COOH = oCH3COONa + oHCl - oNaCl
= = 91 + 425.9 – 126.4 = 390.5
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