Calculating The Degree of ionization: The molar conductance of an infinitely dilute solution `NH_(4)Cl` is `150 S cm^(2) mol^(-1)` and the limiting ionic conductacnes of `Cl^(-)` and `OH^(-)` ions are `76` and `198 S cm^(2) mol^(-1)` respectively. If the molar conductivity of a `0.01 M` solution of `NH_(4)OH` is ` 9.6 S cm^(2) mol^(-1)`, what will be its degree of ionization Strategy: At any concentration `C`, if `alpha` is the degree of ionization then it can be approximented to the ratio of molar conductivity `Lambda_(m)^(c)` at the concentration `C` to limiting molar conducting, `Lambda_(m)^(0)`. We are given `Lambda_(m)^(c)` but we need to find `Lambda_(m)^(0)`.

Calculating The Degree of ionization: The molar conductance of an infi

1.	Calculating The Degree of ionization: The molar conductance of an infinitely dilute solution `NH_(4)Cl` is `150 S cm^(2) mol^(-1)` and the limiting ionic conductacnes of `Cl^(-)` and `OH^(-)` ions are `76` and `198 S cm^(2) mol^(-1)` respectively. If the molar conductivity of a `0.01 M` solution of `NH_(4)OH` is ` 9.6 S cm^(2) mol^(-1)`, what will be its degree of ionization Strategy: At any concentration `C`, if `alpha` is the degree of ionization then it can be approximented to the ratio of molar conductivity `Lambda_(m)^(c)` at the concentration `C` to limiting molar conducting, `Lambda_(m)^(0)`. We are given `Lambda_(m)^(c)` but we need to find `Lambda_(m)^(0)`.
Answer» According to Kohlrausch law `Lamda_(m)^(0)(NH_(4)Cl)=lamda_(NH_(4)^+)^(0)+lamda_(Cl^(-))^(0)` Therefore `lamda_(NH_(4)^(+))^(0)=Lamda_(m)^(0)(NH_(4)Cl)-lamda_(Cl^(-))^(0)` `=(150-76)S cm^(2) mol^(-1)` `=74 S cm^(2)mol^(-1)` According to Kohlrausch law `Lamda_(m)^(0)(NH_(4)OH)=lamda_(NH_(4)^(+))^(0)+lamda_(OH^(-))^(0)` `=(74+198)S cm^(2) mol^(-1)` `= 272 S cm^(2) mol^(-1)` According to Equation (3.30) `alpha=(Lamda_(m)^(c))/(Lamda_(0)^(m))` `=(9.6 "S cm"^(2) "mol"^(-1))/(272 "S cm"^(2) "mol"^(-1))` `=0.0353`

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