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KCET · Chemistry · Electrochemistry

The resistance of \(0.01 \mathrm{~m} \mathrm{KCl}\) solution at \(298 \mathrm{~K}\) is \(1500 \Omega\). If the conductivity of \(0.01 \mathrm{~m} \mathrm{KCl}\) solution at \(298 \mathrm{~K}\) is \(0.1466 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}\). The cell constant of the conductivity cell in \(\mathrm{cm}^{-1}\) is

  1. A \(0.219\)
  2. B \(0.291\)
  3. C \(0.301\)
  4. D \(0.194\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.219\)

Step-by-step Solution

Detailed explanation

Given, concentration of \(\mathrm{KCl}\) solution \(=0.01 \mathrm{~m}\) Resistance \((R)\) of \(\mathrm{KCl}\) solution \(=1500 \Omega\)
Conductivity (K) of \(0.01 \mathrm{~m} \mathrm{KCl}\) solution at \(298 \mathrm{~K}\)
\(\begin{aligned}
&=0.1466 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\
&\therefore \text { Cell constant }\left(G^{*}\right)=\kappa R \\
&=0.1466 \times 10^{-3} \times 1500 \\
&=0.219
\end{aligned}\)