MHT CET · Chemistry · Electrochemistry
What is the conductivity of \(0.05 \mathrm{M} \mathrm{BaCl}_2\) solution if its molar conductivity is \(220 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} ?\)
- A \(0.011 \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(0.022 \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(0.033 \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(0.044 \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0.011 \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Lambda & =\frac{1000 \mathrm{k}}{\mathrm{c}} \\ \mathrm{k} & =\frac{\Lambda \mathrm{c}}{1000} \\ \mathrm{k} & =\frac{220 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.05 \mathrm{~mol} \mathrm{~L}^{-1}}{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}} \\ & =0.011 \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}\)
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