Question

If the values of ${\Lambda }_{\infty }$ of $N{H}_{4}Cl,NaOH$ and $NaCl$ are $130,217$ and $109oh{m}^{-1}c{m}^2$ equiv ${}^{-1}$ respectively, the ${\Lambda }_{\infty }$ of $N{H}_{4}OH$ in $oh{m}^{-1}c{m}^2$ equiv ${}^{-1}$ is

• A. 238
• B. 196
• C. 22
• D. 456

Answer 1

Answer: (A)

${\Lambda }_{\infty }N{H}_{4}Cl={\Lambda }_{\infty }N{H}_4^{+}+{\Lambda }_{\infty }C{l}^{-}=130...(i)$

${\Lambda }_{\infty }NaOH={\Lambda }_{\infty }N{a}^{+}+{\Lambda }_{\infty }O{H}^{-}=217...(ii)$

${\Lambda }_{\infty }NaCl={\Lambda }_{\infty }N{a}^{+}+{\Lambda }_{\infty }C{l}^{-}=109...(iii)$

On adding Eq. (i) and (ii) and subtracting the Eq. (iii)

${\Lambda }_{\infty }N{H}_4^{+}+{\Lambda }_{\infty }O{H}^{-}=130+217-109$

${\Lambda }_{\infty }N{H}_{4}OH=347-109$

$=238oh{m}^{-1}c{m}^2$ equiv ${}^{-1}$