for strong electrolytes the values of molar conductivities at infinite dilution are give below:
electrolyte           ^0m (Sm​2mol-1)
BaCl                 280x10-4
NaCl                 126.5x10-4
NaOH              248X10-4
The molar conductance at infinite dilution for Ba(OH)2
is 










 

Bacl2+2NaOH-2Nacl


280+2(248)-2(126.5)

280+496-253.0=523*10-4 
 
  • 48
The molar conductivities of NaOH,NaCl and BaCl2 at infinite dilution are 2⋅481×10−2Sm2mol−1,1⋅265×10−2Sm2mol−1 and 2⋅800×10−2Sm2mol−1 respectively. The molar conductivity of Ba(OH)2 at infinite dilution will be
  • -5
5.232×10^-2
  • -6
Thanks 😇 Just think what to add and subtract to get Ba(OH)2

  • 1
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