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

Options

A. `wedge_(NH_(4)OH)^(@) = wedge_(Ba(OH)_(3))^(@) + wedge_(NH_(4)Cl)^(@) - wedge_(BeCl_(2))^(@)`
B. `wedge_(NH_(4)OH)^(@) = wedge_(BeCl_(2))^(@) + wedge_(NH_(4)Cl)^(@) - wedge_(Be(OH)_(2))^(@)`
C. `wedge_(NH_(4)OH)^(@) = (wedge_(Be(OH)_(2))^(@) + 2A_(NH_(4)Cl)^(@) - wedge_(BeCl_(2))^(@))/2`
D. `wedge_(NH_(3)Cl)^(@) =(2wedge_(NH_(4)Cl)^(@) - wedge_(Ba(OH)_(4))^(@))/2`

Answer

C

Explanation

`wedge_(Na(OH)_(2))^(@) = wedge_(Ba^(2+))^(@) + 2A_(OH^(-))^(@)`
`wedge_(NH_(4)Cl)^(@) = wedge_(NH_(4)^(+))^(@) + wedge_(Cl^(-))^(@)`
After substituting the above in
`wedge_(NH_(4)OH)^(@) =(wedge_(Ba(OH)_(2))^(@) + 2wedge_(NH_(4)Cl)^(@) - wedge_(BeCl_(2))^(@))/2`
We get, `wedge_(NH_(4)OH)^(@) = wedge_(NH_(4)^(+))^(@) + wedge_(OH^(-))^(@)`