A sample of a mixture of and was dissolved in water and treated with excess . The resulting precipit — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
A $0.5 \text{ g}$ sample of a mixture of $NaCl$ and $NaBr$ was dissolved in water and treated with excess $AgNO_3$. The resulting precipitate of $AgCl$ and $AgBr$ weighed $0.97 \text{ g}$. Find the percentage of $NaCl$ in the original mixture.
Answer: $40.5\%$
💡 Solution & Explanation
Let $x$ be mass of $NaCl$. $\text{Mass } NaBr = 0.5 - x$. <br>$\text{moles } AgCl = x/58.5$; $\text{moles } AgBr = (0.5-x)/103$. <br>$\text{Mass Ppt} = (x/58.5) \times 143.5 + [(0.5-x)/103] \times 188 = 0.97$. <br>$2.453x + 1.825(0.5-x) = 0.97 \Rightarrow 0.628x = 0.0575 \Rightarrow x \approx 0.2025 \text{ g}$. <br>$\% = (0.2025/0.5) \times 100 = 40.5\%$.
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