1.

Limiting molar conductivity of NH_(4)OH(i.e.^^_(m)^(@)(NH_(4)OH)is equal to:

Answer»

`^^_(m(NH_(4)Cl))^(@)+^^_(m(NaCl))^(@)-^^_(m(NAOH))^(@)`
`^^_(m(NaOH))^(@)+^^_(m(NaCl))^(@)-^^_(m(NII_(4)Cl))`
`^^_(m(NH_(4)OH)+)^(@) ^^_(m(NH_(4)Cl))^(@)-^^_(m(HCL))^(@)`
`^^_(m(NH_(4)Cl)+)^(@)^^_(m(NaOH))^(@)-^^_(m(NaCl))^(@)`

Solution :`^^_(m(NH_(4)Cl))^(@)=^^_(mNH_(4)^(+))^(@)+^^_(MCL^(-))^(@)`
`^^_(m(NaOH))^(@)=^^_(mNa^(+))^(@)+^^_(m_(OH^(-)))^(@)`
`^^_(m(NaCl))^(@)=^^_(m_(Na^(+)))^(@)+^^_(m_(Cl^(-)))^(@)`
`THEREFORE ^^_(m(NH_(4)^(+)))^(@)+^^_(m(OH^(-)))^(@)`
`=^^_(m(NH_(4)^(+)))^(@)+^^_(m(Cl^(-)))^(@) + ^^_(m(Na^(+)))^(@)`
`+^^_(m(OH^(-)))^(@)[^^_(m(Cl))^(@)+^^_(m(Cl^(-)))]`
`^^_(m(NH_(4)OH))^(@) =^^_(m(NH_(4)Cl))^(@)+^^_(m(NaOH))-^^_(m(NaCl))^(@)`


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