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Limiting molar conductivity of `NH_4 OH` I e., `Lambda_m (NH_4 OH)` is equal to .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» Correct Answer - B ` 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^+) + [email protected] (Cl^(-))` `=Lambda_m^@ (NH_4Cl) = Lambda) _m^@ (NaOH)` ` = Lambda_m^2 (NACl)`. |
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