Question

For a saturated solution of $AgCl$ at ${25}^{O}C$ ,$k=3.4\times {10}^{-6}{ohm}^{-1}{cm}^{-1}$ and that of $H_{2}O\left(l\right)$ used $2.02\times {10}^{-6}{ohm}^{-1}{cm}^{-1}$. ${\lambda }_m^o$ for AgCl is $138{ohm}^{-1}{cm}^{-1}{mol}^{-1}$ then the solubility of AgCl in moles per liter will be?

• A. ${10}^{-5}$
• B. ${10}^{-10}$
• C. ${10}^{-14}$
• D. ${10}^{-16}$

Answer 1

The correct option is B ${10}^{-10}$

Given $T={138}^{\circ }$ c

$K_{\text{sol}}=3.4\times {10}^{-6},K_{H2O}=2.02\times {10}^{-6}$

For a saturated sol" of AgCl af ${25}^{\circ }C$ is

$\begin{array}{c}s=\frac{1000\times (K_{Sol}-K_{H20})}{\delta }\\ S=\frac{1000\times (3\cdot 4-2.02)\times {10}^{-6}}{138}\\ S={10}^{-5}\\ \text{Ksp of}AgCl=S^2=1\times {10}^{-10}\end{array}$

Correct answer$-\underset{–}{B}$