Question: In Experiment 1, 200.0 mL of solution containing 10.0 mg of SO42- (from Na2SO4) were treated with excess BaCl2 solution to precipitate BaSO4 containing some

In Experiment 1, 200.0 mL of solution containing 10.0 mg of SO42- (from Na2SO4) were treated with excess BaCl2 solution to precipitate BaSO4 containing some coprecipitated Cl_. To find out how much coprecipitated Cl_ was present, the precipitate was dissolved in 35 mL of 98 wt% H2SO4 and boiled to liberate HCl, which was removed by bubbling N2 gas through the H2SO4. The HCl/N2 stream was passed into a reagent solution that reacted with Cl_ to give a color that was measured. Ten replicate trials gave values of 7.8, 9.8, 7.8, 7.8, 7.8, 7.8, 13.7, 12.7, 13.7, and 12.7 _mol Cl_. Experiment 2 was identical to the first one, except that the 200.0-mL solution also contained 6.0 g of Cl_ (from NaCl). Ten replicate trials gave 7.8, 10.8, 8.8, 7.8, 6.9, 8.8, 15.7, 12.7, 13.7, and 14.7 _mol Cl_.
(a) Find the mean, standard deviation, and 95% confidence interval for Cl_ in each experiment.
(b) Is there a significant difference between the two experiments? What does your answer mean?
(c) If there were no coprecipitate, what mass of BaSO4 (FM 233.39) would be expected?
(d) If the coprecipitate is BaCl2 (FM 208.23), what is the average mass of precipitate (BaSO4 + BaCl2) in Experiment 1? By what percentage is the mass greater than the mass in part (c)?

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