Question
Equivalent conductance of $$NaCl,\,HCl$$ and $${C_2}{H_5}COONa $$ at infinete dilution are 126.45, 426.16 and $$91\,{\Omega ^{ - 1}}\,c{m^2},$$ respectively. The equivalent conductance of $${C_2}{H_5}COOH$$ is
A.
$$201.28\,{\Omega ^{ - 1}}\,c{m^2}$$
B.
$$390.71\,{\Omega ^{ - 1}}\,c{m^2}$$
C.
$$698.28\,{\Omega ^{ - 1}}\,c{m^2}$$
D.
$$540.48\,{\Omega ^{ - 1}}\,c{m^2}$$
Answer :
$$390.71\,{\Omega ^{ - 1}}\,c{m^2}$$
Solution :
$$\eqalign{
& {\text{By Kohlrausch's}}\,\,{\text{law}} \cr
& {\lambda _\infty }\,{\text{for}}\,NaCl = {\lambda _{Na}} + {\lambda _{C{l^ - }}}\,\,\,...\left( {\text{i}} \right) \cr
& {\lambda _\infty }\,{\text{for}}\,HCl = {\lambda _{{H^ + }}} + {\lambda _{C{l^ - }}}\,\,\,...\left( {{\text{ii}}} \right) \cr} $$
$${\lambda _\infty }\,{\text{for}}\,{C_2}{H_5}COONa = $$ $${\lambda _{N{a^ + }}} + {\lambda _{{C_2}{H_5}CO{O^ - }}}\,\,\,...\left( {{\text{iii}}} \right)$$
So, $${\lambda _\infty }$$ for $${C_2}{H_5}COOH$$ can be obtained by adding Eqs. (ii) and (iii) and then subtracting Eq. (i)
$$\eqalign{
& = {\lambda _\infty }\,{\text{of}}\,{C_2}{H_5}COONa + {\lambda _\infty }\,{\text{of}}\,HCl - {\lambda _\infty }\,{\text{for}}\,NaCl \cr
& = \left( {91 + 426.16 - 126.45} \right){\Omega ^{ - 1}}\,c{m^2} \cr
& = 390.71\,{\Omega ^{ - 1}}c{m^2} \cr} $$