MHT CET · Chemistry · Electrochemistry
What is the molar conductivity of \(0 \cdot 1\) M NaCl if it's conductivity is \(1 \cdot 06 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}\) ?
- A \(1 \cdot 06 \times 10^{2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- B \(1 \cdot 06 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- C \(9 \cdot 4 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- D \(5 \cdot 3 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(1 \cdot 06 \times 10^{2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}=1.06 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}, \mathrm{C}=0.1 \mathrm{M}\)
Molar conductivity, \(\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{C}}\)
\(\Lambda_{\mathrm{m}}=\frac{1.06 \times 10^{-2} \times 1000}{0.1}=1.06 \times 10^{2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Molar conductivity, \(\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{C}}\)
\(\Lambda_{\mathrm{m}}=\frac{1.06 \times 10^{-2} \times 1000}{0.1}=1.06 \times 10^{2} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
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