Question

A conductivity cell shows resistance of 600 ohm. If conductivity of 0.01 M KCl is $0.0015\ \Omega^{-1}\ \text{cm}^{-1}$, what is cell constant?

• A. $0.60\ \text{cm}^{-1}$
• B. $0.45\ \text{cm}^{-1}$
• C. $0.90\ \text{cm}^{-1}$
• D. $0.75\ \text{cm}^{-1}$

Hint:

Cell constant is simply $\text{Conductivity} \times \text{Resistance}$. Multiplying $0.0015$ by 100 gives $0.15$, then multiplying $0.15$ by 6 yields $0.90$ instantly. Always watch out for simple decimal shift shortcuts!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem provides the measured resistance ($R$) of an electrolytic solution inside a specific conductivity cell along with its specific conductivity ($k$, kappa). We are asked to determine the cell constant ($G^*$) of this electrolytic assembly.

Step 2: Key Formula or Approach:
Specific conductivity ($k$) is defined as the product of conductance ($G$) and the cell constant ($G^*$). Since conductance is the mathematical reciprocal of electrical resistance ($G = \frac{1}{R}$), the equation is written as: $$ k = \frac{1}{R} \times G^* \implies G^* = k \times R $$

Step 3: Detailed Explanation:
Let's substitute the given parameters directly into our rearranged cell constant expression:

• Resistance of the solution, $R = 600\ \Omega$

• Specific conductivity, $k = 0.0015\ \Omega^{-1}\ \text{cm}^{-1}$
Calculating the cell constant $G^*$: $$ G^* = 0.0015\ \Omega^{-1}\ \text{cm}^{-1} \times 600\ \Omega $$ $$ G^* = \frac{15}{10000} \times 600 = \frac{15 \times 6}{10} $$ $$ G^* = \frac{90}{10} = 0.90\ \text{cm}^{-1} $$

Step 4: Final Answer: The cell constant of the conductivity cell is $0.90\ \text{cm}^{-1}$, which perfectly matches option (C).