Total volume of sample 400 mL Trial 1 2 3 Final volume of Na2S2O3 mL 12 23 34

Total volume of sample 400 ml trial 1 2 3 final

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Total volume of sample: 400 mL Trial 1 2 3 Final volume of Na2S2O3, mL 1.2 2.3 3.4 Initial volume of Na2S2O3, mL 0 1.2 2.3
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R. J. C. De Guzman / Chemistry 26.1 (2017) P a g e | 5 Net volume of Na2S2O3, mL 1.2 1.1 1.1 DO Content, ppm O2 2.1693 1.9885 1.9885 Average DO Content, ppm O2 2.049 RSD 41.59451613 ppt Confidence Interval 2.048631741 ± 0.211547453 B. Working Equations [ Na 2 S 2 O 3 ] std = ( 0.15 gof KIO 3 214.0 g mol KIO 3 ) ( 6 mol Na 2 S 2 O 3 1 mol KIO 3 ) ( 0.994 ) ( 0.010 Laliquot 0.1 Ltotal volume ) Volumeof Na 2 S 2 O 3 , L 4 molS 2 O 3 2 ¿ 1 molO 2 ¿ ¿ ( LNa 2 S 2 O 3 used ) ( [ Na 2 S 2 O 3 ] std ) ¿ ppmO 2 = ¿ C. Sample Calculations 1. Molarity of the standard Na2S2O3 solution (All Trials) [ Na 2 S 2 O 3 ] std = ( 0.15 gof KIO 3 214.0 g mol KIO 3 ) ( 6 mol Na 2 S 2 O 3 1 mol KIO 3 ) ( 0.994 ) ( 0.010 Laliquot 0.1 Ltotal volume ) 0.037 LNa 2 S 2 O 3 [ Na 2 S 2 O 3 ] std = 0.011298307 M 0.011 M 2. PPM O2 of the water samples (mg/L)
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R. J. C. De Guzman / Chemistry 26.1 (2017) P a g e | 6 4 mmol S 2 O 3 2 ¿ 1 m molO 2 ¿ ¿ ( 32 m gO 2 1 m molO 2 ) ( 0.0012 ) ( 0.011298307 ) ¿ ppmO 2 ,trial 1 = ¿ 4 mmol S 2 O 3 2 ¿ 1 mmol O 2 ¿ ¿ ( 32 mgO 2 1 mmol O 2 ) ( 0.0011 ) ( 0.011298307 ) ¿ ppmO 2 ,trial 1 = ¿ 4 mmol S 2 O 3 2 ¿ 1 mmol O 2 ¿ ¿ ( 32 mg O 2 1 mmolO 2 ) ( 0.0011 ) ( 0.011298307 ) ¿ ppmO 2 ,trial 1 = ¿ Average ppm O 2 = 2.049 ppm 3. Relative Standard Deviation(RSD) and Confidence Interval RSD = Standard Deviation Mean ( 1000 ppt )= 0.085211846 2.048631741 ( 1000 ppt ) = 41.59451613 ppt ConfidenceLimits ( 95% confidencelevel ) = mean ± t ( standard deviation ) n = 2.048631741 ± 4.30 ( 0.085211846 3
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  • Fall '17
  • Sir Jaden Smith
  • R. J. C. De Guzman

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