The molar conductance at infinite dilution of BaCl₂, NaCl and NaOH are respectively 280x10^-4 ,126.5x10^-4 ,248x10^-4 Sm²mol^-1.What is the molar conductance of Ba(OH)₂ at infinite dilution?

523 x 10^-4 Sm²mol^-1

The correct answer is option 1. \(523 \times 10^{-4}\, \ Sm^2mol^{-1}\).

The molar conductance at infinite dilution of the given electrolytes are:

\(\Lambda _{BaCl_2} =  \Lambda _{Ba^{2+}} + 2\Lambda _{Cl^-} = 280 \times 10^{-4}\, \ Sm^2mol^{-1} ----(i)\)

\(\Lambda _{NaCl} =  \Lambda _{Na^+} + \Lambda _{Cl^-} = 126.5 \times 10^{-4}\, \ Sm^2mol^{-1} ----(ii)\)

\(\Lambda _{NaOH} =  \Lambda _{Na^+} + \Lambda _{OH^-} = 248 \times 10^{-4}\, \ Sm^2mol^{-1} ----(iii)\)

Equating equation (i) + 2 × (iii) - 2(ii), we get

\(\Lambda _{BaCl_2} + 2\times \Lambda _{NaOH} - 2\times \Lambda _{NaCl} = \Lambda _{Ba^{2+}} + 2\Lambda _{Cl^-} + 2 \times [\Lambda _{Na^+} + \Lambda _{OH^-}] - 2 \times [\Lambda _{Na^+} + \Lambda _{Cl^-}]\)

\(⇒ 280 \times 10^{-4} + 2[248 \times 10^{-4}] - 2[126.5 \times 10^{-4}] = \Lambda _{Ba^{2+}} + 2\Lambda _{OH^-}\)

\(⇒ \Lambda _{Ba^{2+}} + 2\Lambda _{OH^-} = 523 \times 10^{-4}\)

\(⇒ \Lambda _{Ba(OH)_2} = 523 \times 10^{-4}\, \ Sm^2mol^{-1}\)