1.

Lamda_(m)^(@)(H_(2)O) is equal to ______

Answer»

`Lamda_(m(HCL))^(@)+Lamda_(m(NaOH))^(@)-Lamda_(m(NaCl))^(@)`
`Lamda_(m(HNO_(3)))^(@)+Lamda_(m(NaNO_(3)))^(@)-Lamda_(m(NaOH))^(@)`
`Lamda_(m(HNO_(3)))^(@)+Lamda_(m(NaOH))^(@)-Lamda_(m(NaNO_(3)))^(@)`
`Lamda_(m(NH_(4)OH))^(@)+Lamda_(m(HCl))^(@)-Lamda_(m(NH_(4)Cl))^(@)`

Solution :This problem is based on the concept of Kohlrausch law of independent MIGRATION of IONS : `(lamda_(m)^(@))` is the sum of limiting MOLAR conductivities of cation `lamda_(+)^(@)` and anion `lamda_(-)^(@)`.
`Lamda_(m(H_(2)O))^(@)=lamda_(m(H^(+)))^(@)+lamda_(m(OH^(-)))^(@)`
The value of `lamda_(m(H^(+)))^(@)` is OBTAINED from the `Lamda_(m(HCl))^(@)` and value of `lamda_(m(OH^(-)))^(@)` is obtained from the `Lamda_(m(NaOH))^(@)` but by doing so `lamda_(m(Na^(+)))^(@)` and `lamda_(m(Cl^(-)))^(@)` are added. So to eliminate them `Lamda_(m(NaCl))^(@)` should be subtracted.


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