Question

questions 1 & 2

Caleulations: Mo lanly of ThOf is 0.25M Sampfe cfD ,2 adu 15 ml MaCl d a25 NACH KO.OISL 0.00325 mal NaOk kons vinalold md of NaOH nal HC2H,O M HC a00375 HC7Of = O.125 M HGHy9 O.03 L Sanple Ns md Ot 5m MO0.01KL- acou, de HOan atulo MHABOecoM H o 0.133m HC2H2 O.03L Jini goiba tond I Questions Jond laitial 1. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Use your results and a density of 1.0 g/mL to investigate this claim. L.O 2. Redo the numerical problem in the Prelaboratory Assignment, substituting H2SO, for HCl. Results Molarity of the NaOH solution: 0.25 m HOJA 1. Trial titration +QPORCA 20 0 Final buret reading (mL): Initial buret reading (mL): 5040 Volume of NaOH solution (mL): 12 2. Exact titrations W. Sample No. w i 4 34 35 35 35 anofran 50 Final buret reading (mL) 50 Initial buret reading (mL) 16 2.133 0.125 Volume of NaOH solution (mL) S O.125 0.127 15 15 O.125 airla Concentration of HC,H,O2 (M) Mean concentration (M)

Answer 1

His claim is false if you have done experiment correct and uses correct molarity and volume for NaOH and acetic acid.

This is because, we found that manufacturer vinegar has molarity of 0.833 M where you found 0.127 M be experiment

Mean Photasily acd is a cet mdady of manfa cteverdti Caleatle he clamed a etie has Rial vinegav Gad,lave ou hpetimad m }m) deaty ram molarty olissolired olute moles ISnumbey Oensity m1 Since acetie acid s t loof1000 li tëv 50 m that means olutia has So fm aceie a cd 950 am watev and vicding usegat (60.052) wor molet ihis mas3 in the acicaif 6.833 molea irn O.333 M melatis So 02 MYou erpei meat hare You Colattal da istalse, Cotedt elone e feimud