Electrolyte:	$$KCl$$	$$KN{O_3}$$	$$HCl$$	$$NaOAc$$	$$NaCl$$
$${\Lambda ^\infty }\left( {S\,\,c{m^2}\,\,mo{l^{ - 1}}} \right):$$	149.9	145	426.2	91	126.5

Calculate $$\Lambda _{HOAc}^\infty $$  using appropriate molar conductances of the electrolytes listed above at infinite dilution in $${H_2}O$$  at $${25^ \circ }C$$

Solution :
$$\eqalign{ & \Lambda _{HCl}^\infty = 426.2\,\,\,...\left( {\text{i}} \right) \cr & \Lambda _{AcONa}^\infty = 91.0\,\,\,...\left( {{\text{ii}}} \right) \cr & \Lambda _{NaCl}^\infty = 126.5\,\,\,...\left( {{\text{iii}}} \right) \cr & \Lambda _{AcOH}^\infty = \left( {\text{i}} \right) + \left( {{\text{ii}}} \right) - \left( {{\text{iii}}} \right) \cr & = \left[ {426.2 + 91.0 - 126.5} \right] \cr & = 390.7 \cr} $$