Question

At $25^{\circ} C$, the molar conductance at infinite dilution for the strong electrolytes $NaOH , NaCl$ and $BaCl _{2}$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} Sm ^{2} mol ^{-1}$ respectively. $\Delta_{ m }{ }^{0} Ba ( OH )_{2}$ in $Sm ^{2} mol ^{-1}$ is

• A. $52.4 \times 10^{-4}$
• B. $524 \times 10^{-4}$
• C. $402 \times 10^{-4}$
• D. $262 \times 10^{-4}$

Answer 1

Answer: (B)

$\Delta^{\circ}_{ Na ^{+}}+\Delta_{ OH ^{-}}^{\circ}=248 \times 10^{-4} Sm ^{2} mol ^{-1}$
$\Delta_{ Na ^{+}}^{\circ}+\Delta_{ Cl ^{-}}^{\circ}=126 \times 10^{-4} Sm ^{2} mol ^{-1} $
$\Delta_{ Ba }^{\circ}{ }^{2+}+\Delta{ }_{2 Cl ^{-}}^{\circ}=280 \times 10^{-4} Sm ^{2} mol ^{-1}$
Now,
$\Delta_{ Ba ( OH )_{2}}^{\circ} =\Delta_{ BaCl _{2}}^{\circ}+2 \Delta_{ NaOH }^{\circ}-2 \Delta_{ NaCl }^{\circ} $
$\Delta_{ Ba ( OH )_{2}}^{\circ} =280 \times 10^{-4}+2 \times 248 \times 10^{-4}-2 \times 126 \times 10^{-4}$
$\Delta_{ Ba ( OH )_{2}}^{\circ} =524 \times 10^{-4} Sm ^{2} mol ^{-1}$