Question

${\lambda }_m^{\circ }$ for $N{H}_{4}Cl,NaOH$ and $NaCl$ are $130,248$ and $126.5oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively. The ${\lambda }_m^{\circ }$ of $N{H}_{4}OH$

• A. $251.5$
• B. $244.5$
• C. $130$
• D. $504.5$

Answer 1

Answer: (A)

${\lambda }_{m(N{H}_{4}OH)}^{\circ }={\lambda }_{m(N{H}_{4}Cl)}^{\circ }+{\lambda }_{m(NaOH)}^{\circ }-{\lambda }_{m(NHCl)}^{\circ }$
$=130+248-126.5=251.5oh{m}^{-1}c{m}^{2}mo{l}^{-1}$