MHT CET · Chemistry · Electrochemistry
What is the molar conductivity of \(0.05 \mathrm{M}\) solution of sodium hydroxide, if its conductivity is \(0.0118 \mathrm{~S} \mathrm{~cm}^{-1}\) at \(298 \mathrm{~K}\) ?
- A \(236 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(423 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(354\mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(590 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(236 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{M}=0.05, \mathrm{k}(\text { conductivity })=0.0118 \mathrm{Scm}^{-1} \\ & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & \Lambda_{\mathrm{m}}=\frac{0.0118 \times 1000}{0.05} \\ & \Lambda_{\mathrm{m}}=236 \mathrm{Scm}^2 \cdot \mathrm{mol}^{-1}\end{aligned}\)
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