[" No for "NaCl,HCl" and "NaAc" are "126.4,425.9" and "91.0Scm^(2)" mot "^(1)" respectively.C."],[" Ao for HAc."]

^^_(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.

A^@_m for NaCl, HC1 and NaAc are 126.4, 425.9 and 91.0 S cm^2 mol_-1 respectively. Calculate A^@ for HAc.

Lambda_(m)^(@) for NaCl, HCl and NaAc are 126.4, 425.9 and "91 S.cm"^(2)."mol"^(-1) respectively. Calculate Lambda^(@) for Hac.

lamda_m^0 for NaCl,HCL and NaAc are 126.4,425.9and 91.0Scm^2mol^(-1) respectively. Calculate ^^_m^0 for Hac.

wedge_(m)^(@) for NaCl,HCl , and NaAc are 126.4, 425.9, and 91.0 S cm^(2) mol^(-1) , respectively . Calculate wedge ^(@) for Hac.

wedge_(m)^(@) for NaCl,HCl , and NaAc are 126.4, 425.9, and 91.0 S cm^(2) mol^(-1) , respectively . Calculate wedge ^(@) for HAc.

wedge_(m)^(@) for NaCl,HCl , and NaAc are 126.4, 425.9, and 91.0 S cm^(2) mol^(-1) , respectively . Calculate wedge ^(@) for Hac.

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 -

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