In 500 ml of solution, 2000 ml of solution is added. The mass of precipitate of AgCl formed is () — Ionic Equilibrium Chemistry Question
Question
In 500 ml of $2.5 \times 10^{-5}\text{ M} – \text{AgNO}_3$ solution, 2000 ml of $5.0 \times 10^{-2}\text{ M} – \text{NaCl}$ solution is added. The mass of precipitate of AgCl formed is ($K_{sp} \text{ of AgCl} = 2 \times 10^{-10}, \text{Ag} = 108$)
Answer: B
💡 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 $= 1.25 \times 10^{-5}\text{ moles} \times 143.5\text{ g/mol} \approx 1.79375 \times 10^{-3}\text{ g} = 1.794\text{ mg}$.
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