1.

The limmiting molar conductivites ^^ ^(@) for NaCl_(2)Kbr and KCl are 126,152 and 150 S cm^(2) respectively the ^^ ^(@) for NaBr is

Answer»

`128Scm^(2)"mol"^(-1)`
`302Scm^(2)"mol"^(-1)`
`278S CM^(2)"mol"^(-1)`
`176Scm^(2)"mol"^(-1)`

SOLUTION :`^^_(NaBr)^(@)=^^_(NaCl)^(@)+^^_(KBr)^(@)-^^_(KCL)^(@)`
`=126+152-150=128 S cm^(2)"mol"^(-1)`


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