Question

The resistance of a conductivity cell filled with 0.1 M KCl solution is \(100 \, \Omega\). If the resistance of the same cell when filled with 0.02 M KCl solution is \(520 \, \Omega\), the molar conductivity of 0.02 M solution (in S cm\(^2\) mol\(^{-1}\)) is (Given: conductivity of 0.1 M KCl solution = \(1.29 \, \text{S m}^{-1}\))

• A. 124
• B. 186
• C. 248
• D. 104

Hint:

- Cell constant \(G^ = \kappa \cdot R\), where \(\kappa\) is conductivity and \(R\) is resistance. \(G^\) is constant for a given cell. - First, calculate \(G^\) using the data for the 0.1 M KCl solution. Remember to keep units consistent (cm vs m). \( \kappa (\text{S cm}^{-1}) = \kappa (\text{S m}^{-1}) \times 10^{-2} \). - Then, calculate \( \kappa \) for the 0.02 M solution using \( \kappa_2 = G^/R_2 \). - Molar conductivity \( \Lambda_m = \frac{\kappa \times 1000}{C} \), where \( \kappa \) is in S cm\(^{-1}\) and \(C\) is molarity (mol/L). The resulting units for \( \Lambda_m \) will be S cm\(^2\) mol\(^{-1}\).

Answer 1

The Correct Option is A

Conductivity \( \kappa \) is related to resistance \(R\) and cell constant \( G^ = L/A \) by \( \kappa = \frac{1}{R} \cdot G^ \).
So, Cell Constant \( G^ = \kappa \times R \).
For 0.
1 M KCl solution: Resistance \( R_1 = 100 \, \Omega \).
Conductivity \( \kappa_1 = 1.
29 \, \text{S m}^{-1} \).
Convert conductivity to S cm\(^{-1}\) for consistency with molar conductivity units often used: \( \kappa_1 = 1.
29 \, \text{S m}^{-1} = 1.
29 \, \text{S} \times (100 \, \text{cm})^{-1} = 1.
29 \times 10^{-2} \, \text{S cm}^{-1} = 0.
0129 \, \text{S cm}^{-1} \).
Cell constant \( G^ = \kappa_1 \times R_1 = (0.
0129 \, \text{S cm}^{-1}) \times (100 \, \Omega) = 1.
29 \, \text{cm}^{-1} \).
The cell constant remains the same for the same cell.
For 0.
02 M KCl solution: Resistance \( R_2 = 520 \, \Omega \).
Conductivity \( \kappa_2 = \frac{G^}{R_2} = \frac{1.
29 \, \text{cm}^{-1}}{520 \, \Omega} \).
\[ \kappa_2 = \frac{1.
29}{520} \, \text{S cm}^{-1} \approx 0.
0024807.
.
.
\, \text{S cm}^{-1} \] Molar conductivity \( \Lambda_m \) is given by \( \Lambda_m = \frac{\kappa \times 1000}{C} \), where \( \kappa \) is in S cm\(^{-1}\) and C is the molar concentration in mol L\(^{-1}\) (M).
For the 0.
02 M KCl solution: Concentration \( C_2 = 0.
02 \) M.
\[ \Lambda_{m,2} = \frac{\kappa_2 \times 1000}{C_2} = \frac{(1.
29/520) \text{ S cm}^{-1} \times 1000 \text{ cm}^3\text{L}^{-1}}{0.
02 \text{ mol L}^{-1}} \] \[ \Lambda_{m,2} = \frac{1.
29 \times 1000}{520 \times 0.
02} \, \text{S cm}^2 \text{mol}^{-1} \] \[ \Lambda_{m,2} = \frac{1290}{520 \times 0.
02} = \frac{1290}{10.
4} \] Calculate \( \frac{1290}{10.
4} \): \( \frac{1290}{10.
4} = \frac{12900}{104} \).
\( 12900 \div 104 \): \( 104 \times 1 = 104 \).
Remainder \( 129-104 = 25 \).
Bring down 0: 250.
\( 104 \times 2 = 208 \).
Remainder \( 250-208 = 42 \).
Bring down 0: 420.
\( 104 \times 4 = 416 \).
Remainder \( 420-416 = 4 \).
So, \( \frac{12900}{104} \approx 124.
038.
.
.
\) \[ \Lambda_{m,2} \approx 124.
04 \, \text{S cm}^2 \text{mol}^{-1} \] This is approximately 124 S cm\(^2\) mol\(^{-1}\).
This matches option (1).