Moles NaOH 38910 3 42410 3 NA Weight CH 3 COOH 463 504 NA RSD for weight CH 3

Moles naoh 38910 3 42410 3 na weight ch 3 cooh 463

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Moles NaOH 3.89*10-34.24*10-3NA Weight % CH3COOH 4.63% 5.04% NA %RSD for weight % CH3COOH 6.0% Algebraic Equations Mole KHP =_KHP sample x 1 mol/204.22g Sample 1: (0.508g) x 1 mol/204.22g = 2.49*10-3= Mole NaOH because of 1:1 ratio [NaOH] =_(moles of NaOH)/(volume of NaOH used in titration) Sample 1: (2.49*10-3)/(0.0134L) = 1.86*10-1_ Moles NaOH added to 5.00mL vinegar solution =_KHP and NaOH are 1:1 ratio at the endpoint moles KHP = moles of NaOH ■■■■■
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2 Sample 1: 2.49*10-3moles of KHP = 2.49*10-3moles of NaOH_ Weight % CH3COOH =_(weight of CH3COOH/weight of vinegar solution) x 100 Sample 1: ((60.05g/mol) x 4.24*10-3mol) / ((1.01g/mL: density of vinegar) x 5.00mL) x 100 = 0.2546/5.05 x 100 = 5.04%Discussion A solution of NaOH was standardised by titration with KHP. The average concentration of the prepared NaOH solution was determined to be _1.97*10-1_ M. This standardised solution of NaOH was used as the titrant to titrate the amount of CH3COOH in a sample of vinegar which was determined to be a weight % of 4.84%_. Conclusions The average weight % of CH3COOH in the sample of vinegar was _4.84_% with a relative standard deviation of 6.0_weight %. References 1.Reimer, M. et al, Properties of Materials,Laboratory Manual, Chemistry 101, pp. 21-26. (University of Victoria: Victoria, B.C.). Summer 2018. Feedback Summary max. Pre-lab quiz: Are all responses correct?3 Laboratory Notebook: Have all data and observations been recorded?1 Report: Are all sections completed? 1 Participation: Was time used well in lab and was student engaged in the experiment?1 Performance evaluation: Did student come prepared to do the exercise and follow the safety guidelines?1 Total mark 7
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