Ncert Exercises & Solutions

$\Lambda _{m(NH_{4}OH)}^{o}$ 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

Λ_{m}^{°} ({NH}_{4} OH)

can be calculated as follows:

$\begin{array}{l} {Λ^{°}}_{m ({NH}_{4} Cl)} = {Λ^{°}}_{m ({NH}_{4}^{+})} + {\dot{Λ}}^{°}_{m ({Cl}^{-})} \\ Λ_{m (NaOH)}^{°} - Λ_{m ({Na}^{+})}^{°} + {Λ^{°}}_{m ({OH}^{-})} \end{array}$
$\begin{array}{l} {Λ^{°}}_{m (NaCl)} = Λ_{m ({Na}^{+})} ° + {Λ^{°}}_{m ({Cl}^{-})} \\ - \end{array}$ $-$ $-$ $Λ_{m ({NH}_{4} Cl)} + Λ_{m (NaOH)} ° - Λ_{m (NaCl)}$ $° = Λ_{m ({NH}_{4} OH)} °$

Hence, option (2) is correct choice.

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