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Molar conductiveiy [Lambda m] at infinit...

Molar conductiveiy `[Lambda _m]` at infinite dilution of Na CL, HCl and `CH_3 COONa` are `126.4, 425, 9` and `91 0 S cm^2 "mol"^(-1)` respectively, `Lambda_m` for `CH_3COOH` will be ,

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Molar conductivities (Lambda_(m)^(@)) at infinite dilution of NaCl, HCl and CH_(3)COONa arc 126.4, 425.9 and 91.0 S cm^(2) mol^(-1) respectively. Lambda_(m)^(@) for CH_(3)COOH will be

Molar conductivities (Lambda_(m)^(@)) at infinite dilution of NaCl, HCl and CH_(3)COONa arc 126.4, 425.9 and 91.0 S cm^(2) mol^(-1) respectively. Lambda_(m)^(@) for CH_(3)COOH will be

Lambda_(m)^(@) of NaCl, HCl and CH_(3)COONa are 126.4, 425.9 and "91.0 S.cm"^(2)."mol"^(-1) respectively. Lambda_(m)^(@) for CH_(3)COOH will be -

If molar conductivities (Lambda_m^0) at infinite dilution of KBr, HBr and CH_3COOK are 151.6, 427.7 and 114.4 S cm^2 mol ^(-1) respectively then (Lambda_m^0) for CH_3COOH will be

If molar conductivities (Lambda_m^0) at infinite dilution of KBr, HBr and CH_3COOK are 151.6, 427.7 and 114.4 S cm^2 mol ^(-1) respectively then (Lambda_m^0) for CH_3COOH will be

lambda_(m)^(@) for NaCl, HCl and CH_(3)COONa are 126.4,425.9 and "91.0 S cm"^(2)//"mol" respectively. Calculate lambda_(m)^(@)" for "CH_(3)COOH .

^^_(m)^(0) for NaCl, HCl and NaAc are 126.4, 425.0 and 91.0 S cm^(2) mol ^(-1) respectively. Calculate ^^(0) for Hac.