Q23 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry
For strong electrolyte $\Lambda_{\mathrm{m}}$ increases slowly with dilution and can be represented by the equation $\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{Ac}^{1 / 2}$ Molar conductivity values of the solutions of strong electrolyte AB at $18^{\circ} \mathrm{C}$ are given below :
\(\begin{array}{|l|c|c|c|c|} \hline \mathrm{c}\;[\mathrm{mol\,L^{-1}}] & 0.04 & 0.09 & 0.16 & 0.25 \\ \hline \Lambda_{\mathrm{m}}\;[\mathrm{S\,cm^{2}\,mol^{-1}}] & 96.1 & 95.7 & 95.3 & 94.9 \\ \hline \end{array}\)
The value of constant A based on the above data $\left[\right.$ in $\left.\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} /(\mathrm{mol} / \mathrm{L})^{1 / 2}\right]$ unit is $\_\_\_\_$.
\(\begin{array}{|l|c|c|c|c|} \hline \mathrm{c}\;[\mathrm{mol\,L^{-1}}] & 0.04 & 0.09 & 0.16 & 0.25 \\ \hline \Lambda_{\mathrm{m}}\;[\mathrm{S\,cm^{2}\,mol^{-1}}] & 96.1 & 95.7 & 95.3 & 94.9 \\ \hline \end{array}\)
The value of constant A based on the above data $\left[\right.$ in $\left.\mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} /(\mathrm{mol} / \mathrm{L})^{1 / 2}\right]$ unit is $\_\_\_\_$.
Correct answer: 4
Solution
The equation $\Lambda_m = \Lambda_m^0 - Ac^{1/2}$ is linear in $c^{1/2}$.
Taking two data points: (c=0.04, $\sqrt{c}$=0.2, $\Lambda_m$=96.1) and (c=0.25, $\sqrt{c}$=0.5, $\Lambda_m$=94.9).
Slope = $\frac{94.9 - 96.1}{0.5 - 0.2} = \frac{-1.2}{0.3} = -4.0$
Since slope = -A, the constant A = 4 S cm² mol⁻¹/(mol/L)^(1/2)
Taking two data points: (c=0.04, $\sqrt{c}$=0.2, $\Lambda_m$=96.1) and (c=0.25, $\sqrt{c}$=0.5, $\Lambda_m$=94.9).
Slope = $\frac{94.9 - 96.1}{0.5 - 0.2} = \frac{-1.2}{0.3} = -4.0$
Since slope = -A, the constant A = 4 S cm² mol⁻¹/(mol/L)^(1/2)
Hence, the correct answer is 4.