\(\Lambda _{m(NH_{4}OH)}^{o}\) is equal to -
1. \(\Lambda _{m(NH_{4}OH)}^{o} \ + \ \Lambda _{m(NH_{4}Cl)}^{o} \ - \ \Lambda _{m(HCl)}^{o}\)
2. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaOH)}^{o} \ - \ \Lambda _{m(NaCl)}^{o}\)
3. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaCl)}^{o} \ - \ \Lambda _{m(NaOH)}^{o}\)
4. \(\ \Lambda _{m(NaOH)}^{o} \ + \ \Lambda _{m(NaCl)}^{o}\ - \ \Lambda _{m(NH_{4}Cl)}^{o}\)

HINT: Use Kohlrausch's law
Explanation:
The value of Λm°NH4OH can be calculated as follows:
 Λ°mNH4Cl=Λ°mNH4++Λ˙°mCl-Λm(NaOH)°-ΛmNa+°+Λ°mOH-
Λ°m(NaCl)=ΛmNa+°+Λ°mCl--               -             -                                 ΛmNH4Cl+Λm(NaOH)°-Λm(NaCl) °=ΛmNH4OH°
 Hence, option (2) is correct choice