Question

Limiting molar conductivity of $N{H}_{4}OH$ [i.e., ${\Lambda }_m^{\circ }\left(N{H}_{4}OH\right)$ ] is equal to:

• A. ${\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)+{\Lambda }_m^{\circ }\left(N{a}_{4}Cl\right)-{\Lambda }_m^{\circ }(NaOH)$
• B. ${\Lambda }_m^{\circ }(NaOH)+{\Lambda }_m^{\circ }(NaCl)-{\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)$
• C. ${\Lambda }_m^{\circ }\left(N_{4}OH\right)+{\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)-{\Lambda }_m^{\circ }(HCl)$
• D. ${\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)+{\Lambda }_m^{\circ }(NaOH)-{\Lambda }_m^{\circ }(NaCl)$

Answer 1

Answer: (D)

According to Kohlrausch law of independent migration of ions:
${\Lambda }_m^{\circ }\left(N{H}_{4}OH\right)={\Lambda }_m^{\circ }\left(N{H}_4^{+}\right)+{\Lambda }_m^{\circ }\left(O{H}^{-}\right)$
$={\Lambda }_m^{\circ }\left(N{H}_4^{+}\right)+{\Lambda }_m^{\circ }\left(C{l}^{-}\right)+{\Lambda }_m^{\circ }\left(O{H}^{-}\right)$
$={\Lambda }_m^{\circ }\left(N{H}_4^{+}\right)-{\Lambda }_m^{\circ }\left(C{l}^{-}\right)-{\Lambda }_m^{\circ }\left(N{a}^{+}\right)$
$={\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)+{\Lambda }_m^{\circ }(NaOH)-{\Lambda }_m^{\circ }(NaCl)$