WBJEE 2026 Physics and Chemistry Paper

Concept:The precipitation reaction replaces $\mathrm{Cl^-}$ ions by $\mathrm{NO_3^-}$ ions.Since their ionic conductances are nearly equal, $\lambda_m$ remains almost constant before the equivalence point.Explanation:The reaction involved is:$< br>\mathrm{AgNO_3 + KCl \rightarrow AgCl \downarrow + KNO_3} < br>$Before the equivalence point:Each added $\mathrm{Ag^+}$ precipitates one $\mathrm{Cl^-}$ as $\mathrm{AgCl}$.In solution, $\mathrm{Cl^-}$ is replaced by $\mathrm{NO_3^-}$.Because $\lambda_{\mathrm{Cl^-}} \approx \lambda_{\mathrm{NO_3^-}}$, the molar conductivity of the solution stays nearly constant.At the equivalence point, all $\mathrm{KCl}$ is converted to $\mathrm{KNO_3}$.After the equivalence point:Excess $\mathrm{AgNO_3}$ is present in the solution.The molar conductivity of $\mathrm{AgNO_3}$ is lower than that of $\mathrm{KNO_3}$.Hence, $\lambda_m$ gradually decreases after the equivalence point.Answer:The correct plot is the one showing $\lambda_m$ almost constant up to the equivalence point and then decreasing as more $\mathrm{AgNO_3}$ solution is added.