Question

The resistance of a conductivity cell filled with 0.02 M KCl solution is 85 \(\Omega\) at \(25^{\circ}C\). Conductivity of this solution is \(0.3 \text{ S m}^{-1}\). Resistance of 0.0025 M \(K_2SO_4\) solution taken in the same cell is 300 \(\Omega\). The molar conductivity of 0.0025 M \(K_2SO_4\) solution (in S \(\text{m}^2 \text{mol}^{-1}\)) is:

• A. \(6.8 \times 10^{-3}\)
• B. \(2.4 \times 10^{-2}\)
• C. \(3.4 \times 10^{-2}\)
• D. \(3.4 \times 10^{-3}\)

Hint:

Remember the conversion factor: \(1 \text{ M} = 1000 \text{ mol/m}^3\).

Answer 1

The Correct Option is C

Step 1: Use the same cell idea.
The same conductivity cell is used for both solutions, so the cell constant $G^{*}$ (which is length over area) is fixed. We will first find it from the known KCl data and reuse it for the sulfate.
Step 2: Find the cell constant from KCl.
The cell constant relates conductivity and resistance through $G^{*} = \kappa \times R$. For KCl, \[ G^{*} = 0.3 \times 85 = 25.5 \text{ m}^{-1} \]
Step 3: Find the conductivity of the sulfate solution.
For the same cell, $\kappa = \dfrac{G^{*}}{R}$. For $K_2SO_4$ with $R = 300\ \Omega$, \[ \kappa = \frac{25.5}{300} = 0.085 \text{ S m}^{-1} \]
Step 4: Convert the concentration into SI units.
The molar conductivity formula needs concentration in mol per cubic metre. $0.0025$ mol/L $= 0.0025 \times 1000 = 2.5$ mol/m$^3$.
Step 5: Apply the molar conductivity formula.
\[ \Lambda_m = \frac{\kappa}{C} = \frac{0.085}{2.5} = 0.034 \text{ S m}^2 \text{ mol}^{-1} \]
Step 6: Express in the option format.
Writing $0.034$ in scientific form gives $3.4 \times 10^{-2}$ S m$^2$ mol$^{-1}$.
\[ \boxed{3.4 \times 10^{-2}} \]