MHT CET · Chemistry · Electrochemistry
Calculate the molar conductivity of \(\mathrm{CH}_2 \mathrm{ClCOOH}\) at zero concentration if molar conductivities of \(\mathrm{HCl}, \mathrm{KCl}\) and \(\mathrm{CH}_2 \mathrm{ClCOOK}\) at zero concentration are \(4.2,1.4\) and \(1.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) respectively.
- A \(3.9 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(4.5 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(6.6 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(1.7 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(3.9 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Lambda_{\mathrm{CH}_2 \mathrm{ClCOOH}}^{\circ} & =\Lambda_{\mathrm{CH}_2 \mathrm{ClCOOK}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{KCl}}^{\circ} \\ \Lambda_{\mathrm{CH}_2 \mathrm{ClCOOH}}^{\circ} & =1.1+4.2-1.4 \\ & =3.9 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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