0.12g of rock salt (NaCl and impurities) was dissolved in water and titrated wit

Question

0.12g of rock salt (NaCl and impurities) was dissolved in water and titrated with 0.1 M silver nitrate. 19.7 mL was required to titrate all the chloride ion. How many moles of sodium chloride were in the original sample and what percent of the sample are impurities?

Explanation / Answer

         The reactionhappens:        NaCl + AgNO3 ->NaNO3 +AgCl(s)       Mole of AgNO3 needs totitrate: 0.1M * 0.0197L =1.97*10-3(mol)       Since ratio mole between NaCl andAgNO3 is 1:1    => Mole of NaCl reacts:1.97*10-3mol         Mass of NaCl in rocksalt: m =Mn = 58.44g/mol * 1.97*10-3mol = 0.115g         Mass of impurities:0.12-0.115 = 0.005g      % of impurities: 0.005/0.12 *100=4.17%

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