In 500 ml of solution, 2000 ml of solution is added. The mass of precipitate of AgCl formed is () — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Moles of $\text{Ag}^+ = 0.5 \times 2.5 \times 10^{-5} = 1.25 \times 10^{-5}\text{ moles}$. Since NaCl is in massive excess, virtually all $\text{Ag}^+$ precipitates as AgCl. Molar mass of AgCl $= 108 + 35.5 = 143.5\text{ g/mol}$. Mass of AgCl precipitated $= 1.25 \times 10^{-5}\text{ moles} \times 143.5\text{ g/mol} \approx 1.79375 \times 10^{-3}\text{ g} = 1.794\text{ mg}$.