MHT CET · Chemistry · Electrochemistry
What is the molar conductivity at infinite dilution of \(\mathrm{CaCl}_{2}\), if the molar conductivity of \(\mathrm{Ca}^{2+}\) ion and \(\mathrm{Cl}^{-}\) ion at infinite dilution is 119 and \(71 \Omega^{-1} \mathrm{~cm}^{2}\) \(\mathrm{mol}^{-1}\)
- A \(431 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- B \(341 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- C \(261 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- D \(126 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(261 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
By Kohlrausch Law of independent migration of ions,
\(
\begin{aligned}
\wedge_{0\left(\mathrm{CaCl}_{2}\right)} &=\lambda_{\mathrm{Ca}^{2+}}^{0}+2 \lambda_{\mathrm{Cl}^{-}}^{0} \\
&=119+2(71) \\
&=261.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
\wedge_{0\left(\mathrm{CaCl}_{2}\right)} &=\lambda_{\mathrm{Ca}^{2+}}^{0}+2 \lambda_{\mathrm{Cl}^{-}}^{0} \\
&=119+2(71) \\
&=261.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
\end{aligned}
\)
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