Given the molar conductance of sodium butyrate, sodium chloride and hydrogen chloride as 83, 127 and $426 \mathrm{~mho} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$ respectively. Calculate the molar conductance of butyric acid at infinite dilution.

JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Given the molar conductance of sodium butyrate, sodium chloride and hydrogen chloride as 83, 127 and

426 \mathrm{~mho} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

25^{\circ} \mathrm{C}

respectively. Calculate the molar conductance of butyric acid at infinite dilution.

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Correct option: A

382

Option A (382) is correct because the molar conductance of butyric acid at infinite dilution is calculated using Kohlrausch's law.

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25^{\circ} \mathrm{C}

0.295 \mathrm{~V}

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

\mathrm{CuS}{\mathrm{O}_4}

{AlCl}_3

STP

({F}=96500 {C})

2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-}

\mathrm{pH}=3

0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S},

\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}}

\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48},

E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V},

\log 2=0.3,

2.303 R T / F=0.06