Chapter 1: Problem 103
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) \(65 \%\)
Short Answer
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Key Concepts
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