MHT CET · Chemistry · Electrochemistry
The conductivity of \(0.04 \mathrm{M} \mathrm{BaCl}_2\) solution is \(0.0112 \Omega^{-1} \mathrm{~cm}^{-1}\) at \(25^{\circ} \mathrm{C}\). what is it's molar conductivity?
- A \(357.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(140.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(44.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(280.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(280.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(
\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}}
\)
Conductivity \((\mathrm{k})=0.0112 \Omega^{-1} \mathrm{~cm}^{-1}\)
\(
\begin{aligned}
& \Lambda_{\mathrm{m}}=\frac{0.0112 \times 1000}{0.04} \\
& =280 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
\)
\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}}
\)
Conductivity \((\mathrm{k})=0.0112 \Omega^{-1} \mathrm{~cm}^{-1}\)
\(
\begin{aligned}
& \Lambda_{\mathrm{m}}=\frac{0.0112 \times 1000}{0.04} \\
& =280 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
\)
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