MHT CET · Chemistry · Electrochemistry
Which is the conductivity of \(0.02 \mathrm{M} \mathrm{HCl}\) solution if molar conductivity of the solution at \(25^{\circ} \mathrm{C}\) is \(412.3 \mathrm{C} \Omega^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\) ?
- A \(8.880 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(8.414 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(8.624 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(8.246 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(8.246 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & 412.3=\frac{\mathrm{k} \times 1000}{0.02} \\ & \mathrm{k}=\frac{412.3 \times 0.02}{1000} \\ & \mathrm{k}=8.246 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}\)
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