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Question
\[\ce{Λ^0_m}_{(NH_4OH)}\] is equal to ______.
Options
\[\ce{Λ^0_m}_{(NH_4OH)} + \ce{Λ^0_m_{(NH_4Cl) } - \ce{Λ^0}_{(HCl)}}\]
\[\ce{Λ^0_m}_{(NH_4Cl)} + \ce{Λ^0_m_{(NaOH) } - \ce{Λ^0}_{(NaCl)}}\]
\[\ce{Λ^0_m}_{(NH_4Cl)} + \ce{Λ^0_m_{(NaCl) } - \ce{Λ^0}_{NaOH)}}\]
\[\ce{Λ^0_m}_{(NaOH)} + \ce{Λ^0_m_{(NACl) } - \ce{Λ^0}_{(NH_4Cl)}}\]
Solution
\[\ce{Λ^0_m}_{(NH_4Cl)} + \ce{Λ^0_m_{(NaOH) } - \ce{Λ^0}_{(NaCl)}}\]
Explanation:
(i) \[\ce{NH4CI ⇌ NH^{+}4 + Cl-}\]
(ii) \[\ce{NaCI ⇌ Na+ + Cl-}\]
(iii) \[\ce{NaOH ⇌ Na+ + OH-}\]
(iv) \[\ce{NH4OH ⇌ NH^{+}4 + OH-}\]
To get equation (iv),
\[\ce{Λ^0_m}_{(NH_4Cl)} + \ce{Λ^0_m_{(NaOH) } - \ce{Λ^0}_{(NaCl)} = \ce{Λ^0_m}_{(NH_4OH)}}\]
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