At 25^o C,the molar conductivities at infinit | vedclass

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(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $HOAc$ (acetic acid) is given by: $

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$NaCl$

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(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $HOAc$ (acetic acid) is given by: $\wedge^o_{HOAc} = \wedge^o_{H^+} + \wedge^o_{OAc^-}$ We are given: $\wedge^o_{NaOAc} = \wedge^o_{Na^+} + \wedge^o_{OAc^-} = 91.0 \, S \, cm^2/mol$ $\wedge^o_{HCl} = \wedge^o_{H^+} + \wedge^o_{Cl^-} = 426.2 \, S \, cm^2/mol$ To obtain $\wedge^o_{HOAc}$,we perform the operation: $\wedge^o_{HOAc} = \wedge^o_{NaOAc} + \wedge^o_{HCl} - \wedge^o_{NaCl}$ $\wedge^o_{HOAc} = (\wedge^o_{Na^+} + \wedge^o_{OAc^-}) + (\wedge^o_{H^+} + \wedge^o_{Cl^-}) - (\wedge^o_{Na^+} + \wedge^o_{Cl^-})$ $\wedge^o_{HOAc} = \wedge^o_{H^+} + \wedge^o_{OAc^-}$ Thus,we need to subtract the molar conductivity of $NaCl$ to cancel out the unwanted ions ($Na^+$ and $Cl^-$). The question asks for the value to be added to the sum of $\wedge^o_{NaOAc}$ and $\wedge^o_{HCl}$ to get the result,which is $-\wedge^o_{NaCl}$. However,among the options provided,$NaCl$ is the substance whose conductivity must be subtracted. If the question implies which substance is involved in the calculation,$NaCl$ is the correct choice.

$c / mol \ L^{-1}$	$\Lambda_m / S \ cm^2 \ mol^{-1}$
$0.000198$	$148.61$
$0.000309$	$148.29$
$0.000521$	$147.81$
$0.000989$	$147.09$

At $25^o C$,the molar conductivities at infinite dilution of $NaOAc$ and $HCl$ in water are $91.0 \, S \, cm^2/mol$ and $426.2 \, S \, cm^2/mol$,respectively. What is the value to be added to calculate the molar conductivity of $HOAc$ at infinite dilution?

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