JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
A \(KCl\) solution of conductivity \(0.14\, S m ^{-1}\) shows a resistance of \(4.19 \,\Omega\) in a conductivity cell. If the same cell is filled with an \(HCl\) solution, the resistance drops to \(1.03 \,\Omega\). The conductivity of the \(HCl\) solution is \(....... \,\times 10^{-2} \,S m ^{-1}\). (Round off to the Nearest Integer).
- A \(28\)
- B \(107\)
- C \(67\)
- D \(57\)
Answer & Solution
Correct Answer
(D) \(57\)
Step-by-step Solution
Detailed explanation
\(\kappa=\frac{1}{R} \cdot G^{*}\) For same conductivity cell, \(G ^{*}\) is constant and hence \(\kappa . R .=\) constant. \(\therefore 0.14 \times 4.19=\kappa \times 1.03\) or,\(\kappa\) of HCl solution \(=\frac{0.14 \times 4.19}{1.03}\) \(=0.5695\, Sm ^{-1}\)…
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