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Question: \(\Delta_{m(NH_{4}OH)}^{0}\)is equal to...

Δm(NH4OH)0\Delta_{m(NH_{4}OH)}^{0}is equal to

A

Δºm(NH4OH)+Δºm(NH4Cl)−Δºm(HCl)\Delta º_{m(NH_{4}OH)} + \Delta º_{m(NH_{4}Cl)} - \Delta º_{m(HCl)}

B

Δºm(NH4Cl)+Δºm(NaOH)−Δºm(NaCl)\Delta º_{m(NH_{4}Cl)} + \Delta º_{m(NaOH)} - \Delta º_{m(NaCl)}

C

Δºm(NH4Cl)+Δºm(NaCl)−Δºm(NaOH)\Delta º_{m(NH_{4}Cl)} + \Delta º_{m(NaCl)} - \Delta º_{m(NaOH)}

D

Δºm(NaOH)+Δºm(NaCl)−Δºm(NH4Cl)\Delta º_{m(NaOH)} + \Delta º_{m(NaCl)} - \Delta º_{m(NH_{4}Cl)}

Answer

Δºm(NH4Cl)+Δºm(NaOH)−Δºm(NaCl)\Delta º_{m(NH_{4}Cl)} + \Delta º_{m(NaOH)} - \Delta º_{m(NaCl)}

Explanation

Solution

Δºm(NH4Cl)+Δºm(NaOH)−Δºm(NaCl)\Delta º_{m(NH_{4}Cl)} + \Delta º_{m(NaOH)} - \Delta º_{m(NaCl)}

}{- \Delta º_{m(Cl^{-})}}$$ $$= \Delta º_{m(Na_{4}^{+})} + \Delta º_{m(OH^{-})} = \Delta º_{m(NH_{4}OH)}$$