Question 21 of 29intermediate🔧 ApplyNumerical2 marks

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146×100.146 \times 10⁻³ S cm⁻¹.

Correct Answer

The cell constant is 0.219 cm⁻¹.

Exercise: EXERCISES | Q: 2.9 | (Page Page 29)
For More Understanding

Explanation

This question tests the fundamental relationship between cell constant, resistance, and conductivity. The cell constant (G*) is defined as the ratio l/A and can be calculated using the formula G* = R × κ, where R is the measured resistance and κ is the known conductivity of the solution. The textbook explicitly states this relationship in equation 2.18 and demonstrates its application in Example 2.8 where a similar calculation is shown.

Solution Steps

  1. Given: Resistance (R) = 1500 Ω, Conductivity (κ) = 0.146×100.146 \times 10⁻³ S cm⁻¹

  2. Formula: Cell constant (G*) = R × κ

  3. Calculation: G* = 1500 Ω × 0.146×100.146 \times 10⁻³ S cm⁻¹ = 1500×0.0001461500 \times 0.000146 cm⁻¹ = 0.219 cm⁻¹

  4. Answer: Cell constant = 0.219 cm⁻¹

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