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

Options

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

C

Explanation

`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^-)^@`