1) At 298 K , the conductivity of a saturated solution , of AgCl in water is 2.6 x 10-6S cm-1. Its solubility product at 298 K, Given : $\lambda^{\infty}(Ag^{+})=63.0 S cm^{2}mol^{-1}$ $\lambda^{\infty}(Cl^{-})=67.0 S cm^{2}mol^{-1}$ A) $2.0\times 10^{-5}M^{2}$ B) $4.0\times 10^{-10}M^{2}$ C) $4.0\times 10^{-16}M^{2}$ D) $2.0\times 10^{-8}M^{2}$ Answer: Option BExplanation: Solubility $S= \frac{1000k}{\lambda_{AgCl}^{0}}=\frac{1000\times2.6\times 10^{-6}}{\lambda_{Ag+}^{0}+\lambda_{Cl-}^{0}}$ $\frac{2.6\times10^{-3}}{63+67}=2\times10^{-5}mol L^{-1}; K_{sp}=S^{2}$
