1. Tardigrade
  2. Question
  3. Chemistry
  4. The molar conductivities of NaOH, NaCl and BaCl2 at infinite dilution are 2.481 × 10-2 S m2 mol-1, 1.265 × 10-2 S m2 mol-1 and 2.800 × 10-2 S m2 mol-1 respectively. The molar conductivity of Ba(OH)2 at infinite dilution will be

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Q. The molar conductivities of $NaOH, NaCl$ and $ BaCl_2$ 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

COMEDKCOMEDK 2014Electrochemistry

Solution:

$\Lambda ^\circ_m [Ba(OH)2] $
$= \Lambda ^\circ_m (BaCl_2) + 2\Lambda ^\circ_m (NaOH) - 2\Lambda ^\circ_m (NaCl)$
= $ (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}$
= $5.232 \times 10^{-2} \, S\, m^2 \, mol^{-1}$