Question
The impure $6 \mathrm{~g}$ of $\mathrm{NaCl}$ is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be $14 \mathrm{~g}$. The \% purity of $\mathrm{NaCl}$ solution would be:(a) $95 \%$(b) $85 \%$(c) $75 \%$(d) $65 \%$
Step 1
Step 1: The reaction that occurs when sodium chloride (NaCl) is treated with silver nitrate (AgNO3) is as follows: \[NaCl + AgNO3 \rightarrow AgCl + NaNO3\] From this reaction, we can see that one mole of NaCl produces one mole of AgCl. Show more…
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