Question

What is the conductivity of 0.02 M KCl solution if cell constant is $1.29\ \text{cm}^{-1}$ with resistance 645 $\Omega$?

• A. $5.0 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• B. $2.0 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• C. $8.3 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• D. $2.5 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$

Hint:

Molar concentration ($0.02\ \text{M}$) is extra information provided in the question stem to test your conceptual clarity; it is not required unless you are asked to compute the molar conductivity ($\Lambda_m$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to find the electrolytic conductivity ($\kappa$) of a potassium chloride solution given its resistance ($R$) and the cell constant ($b$) of the conductivity cell.

Step 2: Key Formula or Approach:
Conductivity ($\kappa$) is defined as the product of conductance ($\frac{1}{R}$) and the cell constant ($b = \frac{l}{A}$): $$\kappa = \frac{\text{Cell Constant}}{Resistance} = \frac{b}{R}$$

Step 3: Detailed Explanation:
Given values: Cell constant, $b = 1.29\ \text{cm}^{-1}$ Resistance, $R = 645\ \Omega$ Substitute these values into the conductivity expression: $$\kappa = \frac{1.29}{645}$$ To make the division straightforward, rewrite $1.29$ in scientific notation as $129 \times 10^{-2}$: $$\kappa = \frac{129 \times 10^{-2}}{645}$$ Notice that $645$ is exactly equal to $5 \times 129$: $$\kappa = \frac{1}{5} \times 10^{-2} = 0.2 \times 10^{-2} = 2.0 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$$

Step 4: Final Answer:
The conductivity of the solution is $2.0 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$, corresponding to option (B).