MHT CET · Chemistry · Electrochemistry
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?
- A \(1.9 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(4.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(2.7 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(3.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(3.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \wedge \mathrm{CH}_2 \mathrm{CICOOH}=\wedge_{\mathrm{CH}_2 \mathrm{CICOOK}}^0+\wedge_{\mathrm{HCI}}^0-\wedge_{\mathrm{KCI}}^0 \\ & =4.2+1.1-1.5 \\ & =3.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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