Question

Molar conductivity of $N{H}_{4}OH$ can be calculated by the equation,

• A. ${\Lambda }_{N{H}_{4}OH}^{\circ }={\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }+{\Lambda }_{N{H}_{4}Cl}^{\circ }-{\Lambda }_{BaC{l}_2}^{\circ }$
• B. ${\Lambda }_{N{H}_{4}OH}^{\circ }={\Lambda }_{BaC{l}_2}^{\circ }+{\Lambda }_{N{H}_{4}Cl}^{\circ }-{\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }$
• C. ${\Lambda }_{N{H}_{4}OH}^{\circ }=\frac{{\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }+2{\Lambda }_{N{H}_{4}Cl}^{\circ }-{\Lambda }_{BaC{l}_2}^{\circ }}{2}$
• D. ${\Lambda }_{N{H}_{4}OH}^{\circ }=\frac{{\Lambda }_{N{H}_{4}Cl}^{\circ }+{\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }}{2}$

Answer 1

Answer: (C)

${\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }={\Lambda }_{B{a}^{2+}}^{\circ }+2{\Lambda }_{O{H}^{-}}^{\circ }$
${\Lambda }_{BaC{l}_2}^{\circ }={\Lambda }_{B{a}^{2+}}^{\circ }+2{\Lambda }_{C{l}^{-}}^{\circ }$
${\Lambda }_{N{H}_{4}Cl}^{\circ }={\Lambda }_{N{H}_4^{+}}^{\circ }+{\Lambda }_{C{l}^{-}}^{\circ }$
After substituting the above in
${\Lambda }_{N{H}_{4}OH}^{\circ }=\frac{{\Lambda }_{Ba{\left(OH\right)}_2}^{\circ }+2{\Lambda }_{N{H}_{4}Cl}^{\circ }-{\Delta }_{BaC{l}_2}^{\circ }}{2}$
we get, ${\Lambda }_{N{H}_{4}OH}^{\circ }={\Lambda }_{N{H}_4^{+}}^{\circ }+{\Lambda }_{O{H}^{-}}^{\circ }$