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Calculate $\wedge_0$ of $\mathrm{CH}_2 \mathrm{ClCOOH}$ if $\wedge_0$ for $\mathrm{HCl}, \mathrm{KCl}$ and $\mathrm{CH}_2 \mathrm{ClCOOK}$ are 4.2, 1.5 and $1.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively?
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The correct answer is:
$3.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\begin{aligned} & \wedge \mathrm{CH}_2 \mathrm{CICOOH}=\wedge_{\mathrm{CH}_2 \mathrm{CICOOK}}^0+\wedge_{\mathrm{HCI}}^0-\wedge_{\mathrm{KCI}}^0 \\ & =4.2+1.1-1.5 \\ & =3.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
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