MHT CET · Chemistry · Electrochemistry
Resistance and conductivity of a cell containing 0.1 M KCl solution at 298 K are 115 ohm and \(1.90 \times 10^{-6} \mathrm{~S} \mathrm{~cm}^{-1}\) respectively. What is the value of cell constant?
- A \(0.165 \mathrm{~cm}^{-1}\)
- B \(1.601 \mathrm{~cm}^{-1}\)
- C \(2.185 \mathrm{~cm}^{-1}\)
- D \(0.218 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(0.218 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
- Resistance: 115 ohm
- Conductivity: \(1.90 \times 10^{-6} \mathrm{~S} \mathrm{~cm}^{-1}\)
- Formula to find the cell constant \(\left(\frac{l}{a}\right)\) :
\(\frac{l}{a}=k \times R\)
where \(k\) is the conductivity and \(R\) is the resistance.
Given Data:
- \(k=1.90 \times 10^{-6} S^{-1}\)
- \(R=115 \Omega\)
Calculation:
Substituting the given values into the formula:
\(\frac{l}{a}=\left(1.90 \times 10^{-6}\right) \times 115=0.2185 \mathrm{~cm}^{-1}\)
Answer:
The cell constant is approximately \(0.2185 \mathrm{~cm}^{-1}\), which matches with Option 4: \(0.218 \mathrm{~cm}^{-1}\).
- Conductivity: \(1.90 \times 10^{-6} \mathrm{~S} \mathrm{~cm}^{-1}\)
- Formula to find the cell constant \(\left(\frac{l}{a}\right)\) :
\(\frac{l}{a}=k \times R\)
where \(k\) is the conductivity and \(R\) is the resistance.
Given Data:
- \(k=1.90 \times 10^{-6} S^{-1}\)
- \(R=115 \Omega\)
Calculation:
Substituting the given values into the formula:
\(\frac{l}{a}=\left(1.90 \times 10^{-6}\right) \times 115=0.2185 \mathrm{~cm}^{-1}\)
Answer:
The cell constant is approximately \(0.2185 \mathrm{~cm}^{-1}\), which matches with Option 4: \(0.218 \mathrm{~cm}^{-1}\).
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