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QAIPM-2012-Ch-1681: NEET 2012 Chemistry - Class 12 AIPM-2012-Ch-1681 Solution: Limiting molar conductivity of $NH_4OH$ [i.e. \(…

Limiting molar conductivity of

NH_4OH

[i.e.

\Lambda_{m(NH_4OH)}^\circ

] is equal to

\Lambda_{m(NH_4OH)}^\circ + \Lambda_{m(NaCl)}^\circ - \Lambda_{m(NaOH)}^\circ

\Lambda_{m(NaOH)}^\circ + \Lambda_{m(NaCl)}^\circ - \Lambda_{m(NH_4Cl)}^\circ

\Lambda_{m(NH_4OH)}^\circ + \Lambda_{m(NH_4Cl)}^\circ - \Lambda_{m(HCl)}^\circ

\Lambda_{m(NH_4Cl)}^\circ + \Lambda_{m(NaOH)}^\circ - \Lambda_{m(NaCl)}^\circ

A $\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0$ _ m ( N H 4 _ O H ) 0 _ + _ m ( N a Cl ) 0 _ _ _ m ( N a O H ) 0 _
B $\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0$ _ m ( N a O H ) 0 _ + _ m ( N a Cl ) 0 _ _ _ m ( N H 4 _ Cl ) 0 _
C $\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0$ _ m ( N H 4 _ O H ) 0 _ + _ m ( N H 4 _ Cl ) 0 _ _ _ m ( H Cl ) 0 _
D $\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$ _ m ( N H 4 _ Cl ) 0 _ + _ m ( N a O H ) 0 _ _ _ m ( N a Cl ) 0 _

Step-by-Step Verified Solution

According to Kohlrausch�s law, the molar conductivity of NH 4 OH _ m ( N H 4 O H ) 0 = _ m ( N H 4 C l ) 0 + _ m ( N a O H ) 0 _ _ m ( N a C l ) 0