Question

${\wedge }_m^{\circ }\left(N{H}_{4}OH\right)$ is equal to

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

Answer 1

The correct option is B ${\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)+{\wedge }_m^{\circ }\left(NaOH\right)-{\wedge }_m^{\circ }\left(NaCl\right)$

According to kohlrausch’s law
the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contribution of its constituent ions.
So
${\wedge }_m^{\circ }\left(N{H}_{4}OH\right)={\wedge }_m^{\circ }N{H}_4^{+}+{\wedge }_m^{\circ }O{H}^{\ominus }...\left(1\right){\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)={\wedge }_m^{\circ }N{H}_4^{+}+{\wedge }_m^{\circ }C{l}^{\ominus }...\left(2\right){\wedge }_m^{\circ }\left(NaOH\right)={\wedge }_m^{\circ }N{H}_4^{+}+{\wedge }_m^{\circ }O{H}^{\ominus }...\left(3\right){\wedge }_m^{\circ }\left(Nacl\right)={\wedge }_m^{\circ }{N}^{+}+{\wedge }_m^{\circ }C{l}^{\ominus }...\left(4\right)$
Eq (1) = Eq(2) + Eq(3) – Eq(4)
Therefore ,
${\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)+{\wedge }_m^{\circ }\left(NaOH\right)-{\wedge }_m^{\circ }\left(NaCl\right)$