Question

Given the limiting molar conductivity as ${Λ}_m^{∞}(HCl)=425.9{Î\copyright }^{−1}c{m}^{2}mo{l}^{−1}$ ${Λ}_m^{∞}(NaCl)=126.4{Î\copyright }^{−1}c{m}^{2}mo{l}^{−1}$ ${Λ}_m^{∞}(C{H}_{3}COONa)=91{Î\copyright }^{−1}c{m}^{2}mo{l}^{−1}$ The molar conductivity, at infinite dilution, of acetic acid (in ${Î\copyright }^{−1}c{m}^{2}mo{l}^{−1}$ ) will be

• A. 481.5
• B. 390.5
• C. 299.5
• D. 516.9

Answer 1

Answer: (B)

Sum of molar conductivity of reactants = sum of molar conductivity of produces Therefore, for the reaction $C{H}_{3}COOH+NaCl\to C{H}_{3}COONa+HCl$ ${Λ}_m^{0}C{H}_{3}COOH={Λ}_m^{0}C{H}_{3}COONa$ $+{Λ}_m^{0}HCl−{Λ}_m^{0}NaCl$ $=91+425.9−126.4$ $=390.5{Î\copyright }^{−1}c{m}^{2}mo{l}^{−1}$