How do I find the mass of acetic acid in a vinegar sample in grams, with the correct amount of signifcant figures? CHEM 181 Experiment #5Part BCalculate the approximate volume of vinegar to react with 25 mL of 0.1 M NAOH.Trial 1Trial 2Trial 3Mass of flask (g)120.056125.237Mass of flask + vinegar sample(3)Mass of vinegar sample (g)SIE:3と1131.911123.1223.0GCO008LOInitial buret reading of NaOH(m)Final buret reading of NaOH25.2524.952 6.05(m)Volume of NaOH dispensed(mL)[NaOH] (M), average from Part25.2524.95O. 100A.Moles of NaOH dispensedOi 002525 mol 0.00249 Sma0.002cos molMoles of acetic acid in vinegar0.002525 mol 0.00249S mollo.002GOS molsampleMass of acetic acid in vinegarsample (g)Percent by mass of acetic acid invinegar sample (%)Mass SarAverage percent by mass of acetic acid in vinegar sample (%)Standard deviation of percent by mass of acetic acid in vinegar sample (%).

Given figure: It is required to find average percent by mass of acetic acid in vinegar sample and standard deviation of percent by mass acetic acid in vinegar sample (%)During titration the vinegar solution gets neutralized when Number of moles of base (0.1M NaOH solution)=Number of moles of acid (acetic acid in vinegar) Mass of acetic acid in vinegar sample (g)=Number of moles of acetic acid in vinegar*Molecular weight ofacetic acid Molecular weight of acetic acid= 60.052 g/mol Percent by mass of acetic acidin vinegar sample (%) = Mass of acetic acid in vinegar sample (g)mass of vinegar sample (g)*100 From the given table considering mass of vinegar sample and last four columns and solving the above equations Trial 1 Trial 2 Trial 3 mass of vinegar sample (g) 3.066 3.078 3.59 Moles of NaOH dispensed (mol) 0.002525 0.002495 0.002005 Moles of acetic acid in vinegar sample (mol) 0.002525 0.002495 0.002005 Mass of acetic acid in vinegar sample (g) 0.1516313 0.14983 0.120404 Percent by mass of acetic acid in vinegar sample (%) 4.94557404 4.867763 3.353879 From table we can see that the percent by mass of acetic acid of trail 1 and trial 2 are close to each other but the value of trail 3 has more deviation. This means that they may be an error in trail 3 measurements Average percent by mass of acetic acid in vinegar sample (%)=Sum of percent by mass of acetic acid in vinegar sample in each trailNumber of trails Average percent by mass of acetic acid in vinegar sample (%)=4.94557404+4.867763+3.3538793=4.389072 % Standard deviationof mass percent bymass of acetic acidin vinegar sample=∑(ipercent mass of acetic acid in trial i-Average percent by mass of acetic acidNumber of trials -1 Standard deviationof mass percent bymass of acetic acidin vinegar sample=(4.94557404-4.389072)2+(4.867763-4.389072)2+(3.353879-4.389072)23-1=0.805232 %