^ _mH_2O is equal to ....... a. ^ _ m(HCl) + ^ _ m(NaOH) - ^ _ m(NaCl) b. ^ _ m(HNO_3) + ^ _ m(NaNO_3) - ^ _ m(NaOH) c. ^ _ (HNO_3) + ^ _ m(NaOH) - ^ _ m(NaNO_3) d. ^ _ m(NH_4OH) + ^ _ m(HCl) - ^ _ m(NH_4Cl)

^ _mH_2O is equal to ....... a. ^ _ m(HCl) + ^ _ m(NaOH) - ^ _ m(…

MCQ

$\wedge^{{\circ}}_\text{m}\text{H}_2\text{O}$ is equal to $.......$
$a. \wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$b. \wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}-\wedge^{{\circ}}_{\text{m}(\text{NaOH})}$
$c. \wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
$d. \wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}$

A
$a$ and $b$
B
$b$ and $c$
✓
$a$ and $c$
D
$a$ and $d$

✓

Answer

Correct option: C.

$a$ and $c$

$\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O)}}=\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}$
$\wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O})}$
$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(0\text{H}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}$
$\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{0})}$
However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here $\ce{NH_4OH}$ is a weak electrolyte of which complete decomposition is not possible.