Limiting molar conductivity of `NH_4 OH` I e.,
`Lambda_m (NH_4 OH)` is equal to .

Options

A. `Lambda_m (NH_4 OH) + Lambda_m(NH_4Cl)-Lambda_m(HCl)`
B. `Lambda_m (NH_4 OH) + Lambda_m(Na_OH)-Lambda_m(HCl)`
C. `Lambda_m (NH_4 OH) + Lambda_m(NsCl)-Lambda_m(HCl)`
D. `Lambda_m (NH_4 OH) + Lambda_m(NaCl)-Lambda_m(HCl)`

Answer

B

Explanation

` Lambda_m^@ (NH_4 Cl) = Lambda_m^@ (NH_4^+) + Lambda_m^@ (Cl^(-) )` …(i)
` A_m^@ (NAOH) = Lambda_m^@(Na^+) + Lambda_m^2 (OH^(-))` ..(ii)
`Lambda_m^2 (NaCl)= Lambda_m^@ (Na^+) +Lambda_m^@ (Cl^(-))` …(iii)
By (i) + (ii) + (iii)
`Lambda_m^@ (NH_4 Cl) =Lambda_m^2 (NH_4^+) + Lambda_m@ (Cl^(-))`
`=Lambda_m^@ (NH_4Cl) = Lambda) _m^@ (NaOH)`
` = Lambda_m^2 (NACl)`.