JEE Advanced 2026 Paper 2 Chemistry Question 1 Solution

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

$\begin{array}{|c|c|c|} \hline \text{Salt} & \text{Concentration (M)} & \text{Molar conductivity (S cm}^2 \text{ mol}^{-1}\text{)} \\ \hline NaNO_3 & 0.01 & 111 \\ & 0.04 & 101 \\ \hline NaCl & 0.01 & 117 \\ & 0.04 & 107 \\ \hline AgNO_3 & 0.01 & 125 \\ & 0.04 & 116 \\ \hline \end{array}$

The conductivity of a saturated aqueous solution of $AgCl$ is $1.40 \times 10^{-6}$ S cm $^{-1}$ at 300 K. If the solubility of $AgCl$ in water at 300 K is $X$ mol L $^{-1}$ , then $\log_{10}(X^{-1})$ is

(Assume that $AgCl$ dissolved in water ionizes completely and that the molar conductivity of saturated $AgCl$ solution is equal to its limiting molar conductivity.)

Detailed Solution

Using Kohlrausch's law: $\Lambda_m = \Lambda_m^\infty - b\sqrt{C}$

For $NaNO_3$ : $111 = \Lambda_m^\infty - b\sqrt{0.01} = \Lambda_m^\infty - 0.1b$ $101 = \Lambda_m^\infty - b\sqrt{0.04} = \Lambda_m^\infty - 0.2b$ Subtracting equations: $10 = 0.1b \implies b = 100$ . $\Lambda_m^\infty(NaNO_3) = 111 + 10 = 121$ S cm $^2$ mol $^{-1}$ .

For $NaCl$ : $117 = \Lambda_m^\infty - b(0.1)$ $107 = \Lambda_m^\infty - b(0.2)$ $10 = 0.1b \implies b = 100$ . $\Lambda_m^\infty(NaCl) = 117 + 10 = 127$ S cm $^2$ mol $^{-1}$ .

For $AgNO_3$ : $125 = \Lambda_m^\infty - b(0.1)$ $116 = \Lambda_m^\infty - b(0.2)$ $9 = 0.1b \implies b = 90$ . $\Lambda_m^\infty(AgNO_3) = 125 + 9 = 134$ S cm $^2$ mol $^{-1}$ .

Now, $\Lambda_m^\infty(AgCl) = \Lambda_m^\infty(AgNO_3) + \Lambda_m^\infty(NaCl) - \Lambda_m^\infty(NaNO_3)$ $\Lambda_m^\infty(AgCl) = 134 + 127 - 121 = 140$ S cm $^2$ mol $^{-1}$ .

For saturated solution, $\Lambda_m = \Lambda_m^\infty = \frac{1000 \kappa}{X}$ $140 = \frac{1000 \times 1.40 \times 10^{-6}}{X}$ $X = \frac{1.40 \times 10^{-3}}{140} = 10^{-5}$ mol L $^{-1}$ .

Therefore, $\log_{10}(X^{-1}) = \log_{10}(10^5) = 5$ .

Question 1

Detailed Solution

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