Question

The molar conductivities of $ {NaOH, NaCl}$ and $ { BaCl2}$ at infinite dilution are $ {2.481 \times 10^{-2}\, S \, m^2\, mol^{-1}, 1.265 \times 10^{-2} \, S\, m^2\, mol^{-1}}$ and $ {2.800 \times 10^{-2} \, S \, m^2\, mol^{-1}}$ respectively. The molar conductivity of $ {Ba(OH)_2}$ at infinite dilution will be

• A. $\ce{5.232 \times 10^{-2} \, S \, m^2\, mol^{-1}}$
• B. $\ce{9.654 \times 10^{-2} \, S \, m^2\, mol^{-1}}$
• C. $\ce{4.016 \times 10^{-2} \, S \, m^2\, mol^{-1}}$
• D. $\ce{1.145 \times 10^{-2} \, S \, m^2\, mol^{-1}}$

Answer 1

The Correct Option is A

$\bigwedge^\circ_m { [Ba(OH)2]} $
$= \bigwedge^\circ_m { (BaCl_2)} + 2\bigwedge^\circ_m { (NaOH)} - 2\bigwedge^\circ_m {(NaCl)}$
= $ { (2.8 + 2 \times 2.481 - 2 \times 1.265) \times 10^{-2}\, S \, m^2\, mo1^{-1}}$
= $ { (2.8 + 4. 962 - 2.53) \times 10^{-2} \, S\, m^2 \, mol^{-1}}$
= $ {5.232 \times 10^{-2} \, S\, m^2 \, mol^{-1}}$