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2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgCl
NaCl + AgNO3 --> AgCl + NaNO3
1 mole of AgCl = 1 mole of NaCl
The sample contained 0.01426 moles of NaCl
0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl
0.8334 / 0.8421 = 98.97% pure

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Q: An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
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