Question

The resistance of a conductivity cell with cell constant 1.14 cm\(^{-1}\), containing 0.001 M KCl at 298 K is 1500 \(\Omega\). The molar conductivity of 0.001 M KCl solution at 298 K in S cm\(^2\) mol\(^{-1}\) is _________. (Integer answer)

Hint:

Remember the flow of calculation: From Resistance (R) and Cell Constant (G*), find Conductivity (\(\kappa\)). Then from Conductivity (\(\kappa\)) and Concentration (C), find Molar Conductivity (\(\Lambda_m\)). Pay close attention to units to correctly use the factor of 1000.

Answer 1

Correct Answer: 760

Step 1: Understanding the Question:
We are given the resistance, cell constant, and concentration for a solution and asked to calculate its molar conductivity.
Step 2: Key Formulas:
1. Conductivity (\(\kappa\)): It relates the resistance (R) and the cell constant (\(G^* = l/A\)).
\[ \kappa = \frac{1}{R} \times \frac{l}{A} = \frac{G^*}{R} \] 2. Molar Conductivity (\(\Lambda_m\)): It relates conductivity to the molar concentration (C).
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \] (The factor of 1000 is used when \(\kappa\) is in S cm\(^{-1}\) and C is in mol L\(^{-1}\) to get \(\Lambda_m\) in S cm\(^2\) mol\(^{-1}\)).
Step 3: Calculate Conductivity (\(\kappa\)):
Given:
- Cell constant, \(G^* = 1.14\) cm\(^{-1}\).
- Resistance, \(R = 1500 \, \Omega\).
\[ \kappa = \frac{1.14 \, \text{cm}^{-1}}{1500 \, \Omega} = 0.00076 \, \text{S cm}^{-1} \] Step 4: Calculate Molar Conductivity (\(\Lambda_m\)):
Given:
- Concentration, \(C = 0.001\) M.
\[ \Lambda_m = \frac{(0.00076 \, \text{S cm}^{-1}) \times 1000 \, \text{cm}^3\text{/L}}{0.001 \, \text{mol/L}} \] \[ \Lambda_m = \frac{0.76}{0.001} = 760 \, \text{S cm}^2 \text{mol}^{-1} \] Step 5: Final Answer:
The molar conductivity of the solution is 760 S cm\(^2\) mol\(^{-1}\).