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Electrochemistry — Chemistry STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Given that $\Lambda^{0}_{\text{m}}(\text{HCl})=426\text{ S cm}^2\text{mol}^{-1},\Lambda^{0}_{\text{m}}(\text{NaCl})\\=126\text{ S cm}^2\text{mol}^{-1}$
$\Lambda^{0}_{\text{m}}(\text{CH}_3\text{COONa})=91\text{ S cm}^2\text{mol}^{-1}$
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Answer

$\Lambda^{0}_{\text{m}}(\text{CH}_3\text{COOH})=\Lambda^{0}_{\text{CH}_3\text{COO}^-}+\Lambda^{0}_{\text{H}^+}$
$=\Lambda^{0}_{\text{CH}_3\text{COO}^-}=\Lambda^{0}_{\text{Na}^{+}}+\Lambda^{0}_{\text{H}^{+}}+\Lambda^{0}_{\text{Cl}^{-}}-\Big(\Lambda^{0}_{\text{Na}^{+}}+\Lambda^{0}_{\text{Cl}^{-}}\Big)$
$=\Lambda^{0}_{\text{m(CH}_3\text{COONa})}+\Lambda^{0}_{\text{m(HCl})}-\Lambda^{0}_{\text{m(NaCl})}$
$=(91+426+126)\text{S cm}^2\text{mol}^{-1}$
$=391\text{ S cm}^2\text{mol}^{-1}$

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