The equivalent conductance of an infinitely dilute solution of `NH_(4)Cl` is 150 and te ionic conductance of `OH^(-)` ad `Cl^(-)` ions are 198 and 76

Question

The equivalent conductance of an infinitely dilute solution of `NH_(4)Cl` is 150 and te ionic conductance of `OH^(-)` ad `Cl^(-)` ions are 198 and 76

The equivalent conductance of an infinitely dilute solution of `NH_(4)Cl` is 150 and te ionic conductance of `OH^(-)` ad `Cl^(-)` ions are 198 and 76 respectively. What will be the equivalent conductance of the solution of `NH_(4)OH` at infinite dilution. if the equivalent conductance of a 0.01 N solution `NH_(4)OH` is 9.6, what will be its degree of dissociation?

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - `272, 0.0353`
`^^_(eq)^(@)(NH_(4)Cl)=150=^^_(eq)^(@)(NH_(4)^(+))+^^_(eq)^(@)(Cl^(-))`
`^^_(OH^(-))^(@)=198,^^_(Cl^(-))^(@)=76,^^_(eq)(NH_(4)OH)=9.6,C=0.01`
`^^_(eq)^(@)(NH_(4)OH)=150-76+198=272`
`alpha=(^^_(eq)(NH_(4)OH))/(^^_(eq)^(@)(NH_(4)OH))=(9.6)/(272)=0.0353`

5 views Answered 2023-02-05 15:29