^ _mH_2O is equal to ___________.

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MCQ

$\wedge^{{\circ}}_\text{m}\text{H}_2\text{O}$ is equal to ___________.

A
$\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
B
$\wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}-\wedge^{{\circ}}_{\text{m}(\text{NaOH})}$
C
$\wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
D
$\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}$

✓

Answer

$\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$

$\wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$

Explanation:

$\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O)}}=\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$

$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}$

$\wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O})}$

$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(0\text{H}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}$

$\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{0})}$

However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here NH₄OH is a weak electrolyte of which complete decomposition is not possible.