Calkulate the approximate volume of O.IM NaOHhe volume of the vinegar sample nee

Question

Calkulate the approximate volume of O.IM NaOHhe volume of the vinegar sample needed for the analyses (Part A.1) the approxin, 0.025 L =0.0025 mo! Na01t 0.0025 4 (0.05 o. 15y acet acad 3.0g 3.0m Brand of vinegar or unknown no. C. Trial 1 Trial 2 Trial 1 1. Mass of flask (g) 2 Mass of fask + vinegar () L1L.513 3.Mass ofvinegar (g) B Analysis of Vinegar Sample 1. Buret reading of NaOH, initial (mL) 2 Buret reading of NaOH, final (mL) 3.Volume of NaOH used (nL) 1L5 4. Molar concentration of NaOH (mon) s. Moles of NaOH added (mol) 2.003 6 Moles of CH,COOH in vinegar (mol)- t vinegar (e)513 102.119 as.o3 038 83 4103 2.401 2.152.841 2.1112 Sample yae 0.000 13.2 u.o - 54.I ls.2 3.2 | 25.5 s.S 15.20 0.113 0.113 | 015 nol) 1.3010 130 1302.05- CH,COOH in vinegar (mol)l.3x10-3 2.okx10-3 0.121 Hat, 2.0303 .15xlo-3 2.03x10-311-15xn? 0:122 t.29 0.0S | 7. Mss of CH,COOH in vinegar (e) 0. & Percent by mass ofCH,COOH in vinegar (%) 2.611 3: sun verage percent by mass of CH,COOH in vinegar (%) Calculations ons for Trial 1 of the first vinegar sample on next page. Exper

Explanation / Answer

Trail-1:

3. Mass of vinegar = 102.119 g - 99.262 g = 2.857 g

B.3: Volume of NaOH used = final buret reading - initial buret reading = 18.20 mL - 0.000 mL = 18.20 mL

4: molar concentration of NaOH, M = 0.113 M

5: Moles of NaOH = M x V(L) = 0.113 mol/L x 18.20 mL x (1 L / 1000 mL) = 0.0020566 mol = 2.06x10-3 mol

6: Since 1 mole of CH3COOH reacts with 1 mole of NaOH in the balanced neutralization reaction,

moles of CH3COOH in vinegar = moes of NaOH =  2.06x10-3 mol

7: molecular mass of CH3COOH = 60.05 g/mol

Hence mass of CH3COOH in vinegar = moles of CH3COOH x molar mass = 2.06x10-3 mol x 60.05 g/mol

= 0.124 g

8: Percent by mass of CH3COOH in vinegar = [(mass of CH3COOH) / (mass of vinegar)] x 100

= (0.124 g / 2.857 g) x 100 = 4.34 %

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