Question

For strong electrolyte \(\Lambda_m\) increases slowly with dilution and can be represented by the equation: \[ \Lambda_m = \Lambda_m^0 - A\sqrt{c} \] Molar conductivity values of the solutions of strong electrolyte AB at 18\(^\circ\)C are given below: \[ \begin{array}{|c|c|c|c|c|} \hline c \,[\text{mol L}^{-1}] & 0.04 & 0.09 & 0.16 & 0.25
\hline \Lambda_m \,[\text{S cm}^2 \text{ mol}^{-1}] & 96.1 & 95.7 & 95.3 & 94.9
\hline \end{array} \] The value of constant \(A\) based on the above data [in \(\text{S cm}^2 \text{ mol}^{-1}/(\text{mol L}^{-1})^{1/2}\)] unit is:

Hint:

For strong electrolytes, plot \(\Lambda_m\) vs. \(\sqrt{c}\). The slope gives \(A\), and the intercept at \(\sqrt{c} = 0\) gives \(\Lambda_m^0\).

Answer 1

Correct Answer: 4

Concept: For strong electrolytes, molar conductivity varies with concentration as: \[ \Lambda_m = \Lambda_m^0 - A\sqrt{c} \] This is known as the Debye–Hückel–Onsager equation. The slope of the plot of \(\Lambda_m\) vs. \(\sqrt{c}\) gives the value of \(A\).
Step 1: Tabulate values of \(\sqrt{c}\). \[ \sqrt{0.04} = 0.2, \quad \sqrt{0.09} = 0.3, \quad \sqrt{0.16} = 0.4, \quad \sqrt{0.25} = 0.5 \]
Step 2: Use two data points to calculate slope. Take \(c = 0.04\) and \(c = 0.25\): \[ \Lambda_m(0.04) = 96.1, \quad \Lambda_m(0.25) = 94.9 \] \[ \Delta \Lambda_m = 96.1 - 94.9 = 1.2 \] \[ \Delta \sqrt{c} = 0.5 - 0.2 = 0.3 \] \[ A = \frac{\Delta \Lambda_m}{\Delta \sqrt{c}} = \frac{1.2}{0.3} = 4.0 \]
Step 3: Verify with other points. Between \(c = 0.09\) and \(c = 0.16\): \[ \Delta \Lambda_m = 95.7 - 95.3 = 0.4, \quad \Delta \sqrt{c} = 0.4 - 0.3 = 0.1 \] \[ A = \frac{0.4}{0.1} = 4.0 \]
Step 4: Conclusion. The constant \(A\) is: \[ A = 4.0 \,\text{S cm}^2 \text{ mol}^{-1}/(\text{mol L}^{-1})^{1/2} \]