Molar conductivity of `NH_4OH` can be calculated by the equation.A. `Lambda_(NH_4OH)^@=Lambda_(Ba(OH)_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@`B. `Lambda_(NH_4OH)^@=Lambda_(BaCl_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(Ba(OH)_2)^@`C. `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+ 2Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@)/2`D. `Lambda_(NH^4OH)^@=(Lambda_(NH_4Cl)^@+Lambda_(Ba(OH)_2)^@)/2`

Molar conductivity of `NH_4OH` can be calculated by the equation.A. `L

1.	Molar conductivity of `NH_4OH` can be calculated by the equation.A. `Lambda_(NH_4OH)^@=Lambda_(Ba(OH)_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@`B. `Lambda_(NH_4OH)^@=Lambda_(BaCl_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(Ba(OH)_2)^@`C. `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+ 2Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@)/2`D. `Lambda_(NH^4OH)^@=(Lambda_(NH_4Cl)^@+Lambda_(Ba(OH)_2)^@)/2`
Answer» Correct Answer - C `Lambda_(Ba(OH)_2)^@=Lambda_(Ba^(2+))^@+ 2Lambda_(OH^-)^@` `Lambda_(BaCl_2)^@=Lambda_(Ba^(2+))^@+2Lambda_(Cl^-)^@` `Lambda_(NH_4Cl)^@=Lambda_(NH_4)^+ +Lambda_(Cl^-)^@` After substituting the above in `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+2Lambda_(NH_4Cl)^@ - Lambda_(BaCl_2)^@)/2` we get, `Lambda_(NH_4OH)^@ =Lambda_(NH_4^+)^@+ Lambda_(OH^-)^@`

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