Question

What is the conductivity of 0.01 M NaCl solution if resistance and cell constant of NaCl solution are 375 ohms and $0.5\ \text{cm}^{-1}$ respectively at 298 K ?

• A. $7.50 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• B. $1.333 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• C. $1.333 \times 10^{-4}\ \Omega^{-1}\ \text{cm}^{-1}$
• D. $1.875 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$

Hint:

Conductivity increases when resistance decreases for a given cell constant.

Answer 1

The Correct Option is B

Step 1: Formula for conductivity.
Conductivity ($\kappa$) is given by:
\[ \kappa = \frac{\text{Cell constant}}{\text{Resistance}} \]
Step 2: Substitution of given values.
\[ \kappa = \frac{0.5}{375} \]
Step 3: Calculation.
\[ \kappa = 1.333 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1} \]
Step 4: Conclusion.
Therefore, the conductivity of the solution is $1.333 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$.