Question

What is the resistance of 0.01 M KCl solution if its conductivity is $200\ \mathrm{ohm}^{-1}\ \mathrm{cm}^{-1}$ and cell constant is $1\ \mathrm{cm}^{-1}$?

• A. $1 \times 10^{-3}\ \mathrm{ohm}$
• B. $5 \times 10^{-3}\ \mathrm{ohm}$
• C. $4 \times 10^{-3}\ \mathrm{ohm}$
• D. $2 \times 10^{-3}\ \mathrm{ohm}$

Hint:

The molarity value ($0.01\ \mathrm{M}$) given in the prompt is extra data meant to distract you! When computing resistance directly from a known conductivity and cell constant, the concentration of the solution is already accounted for inside the conductivity value itself.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to determine the electrical resistance ($R$) of a $0.01\ \mathrm{M}$ potassium chloride solution, given its electrolytic conductivity ($\kappa = 200\ \mathrm{ohm}^{-1}\ \mathrm{cm}^{-1}$) and the cell constant ($\frac{l}{A} = 1\ \mathrm{cm}^{-1}$).

Step 2: Key Formula or Approach:
Conductivity ($\kappa$) is mathematically defined as the product of conductance ($G = \frac{1}{R}$) and the cell constant ($\frac{l}{A}$):
$$\kappa = \frac{1}{R} \cdot \left(\frac{l}{A}\right)$$ Rearranging this equation to directly isolate and solve for resistance ($R$) yields:
$$R = \frac{\left(\frac{l}{A}\right)}{\kappa}$$

Step 3: Detailed Explanation:
Extract the numerical values provided by the problem description:
Cell constant, $\frac{l}{A} = 1\ \mathrm{cm}^{-1}$
Conductivity, $\kappa = 200\ \mathrm{ohm}^{-1}\ \mathrm{cm}^{-1}$
Substitute these parameters into our rearranged expression:
$$R = \frac{1\ \mathrm{cm}^{-1}}{200\ \mathrm{ohm}^{-1}\ \mathrm{cm}^{-1}}$$ $$R = \frac{1}{200}\ \mathrm{ohm}$$ $$R = 0.005\ \mathrm{ohm}$$ Converting this decimal value into standard scientific power notation gives:
$$R = 5 \times 10^{-3}\ \mathrm{ohm}$$

Step 4: Final Answer:
The resistance of the solution is $5 \times 10^{-3}\ \mathrm{ohm}$, matching option (B).