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Calculate $\wedge_{H O A c}^{\infty}$ Using appropriate molar conductances of the electrolytes listed above at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ at $25^{\circ} \mathrm{C}$
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The correct answer is:
390.7
390.7
$\wedge_{\mathrm{AcOH}}^{\infty}=\wedge_{\mathrm{HCl}}^{\infty}+\wedge_{\mathrm{AcONa}}^{\infty}-\wedge_{\mathrm{NaCl}}^{\infty}$
$=390.7$
$=390.7$
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