MHT CET · Chemistry · Electrochemistry
Molar conductivity of \(0.04 \mathrm{M} \mathrm{BaCl}_2\) solution is \(230 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\). what is its conductivity?
- A \(2.3 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(9.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(6.9 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(4.6 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(9.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 100}{\mathrm{M}} \\ & 230=\frac{\mathrm{k} \times 100}{0.04} \\ & \mathrm{k}=\frac{230 \times 0.04}{1000} \\ & =9.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}\)
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