Question

Date Instructor Fill in the table below. Record your data with the correct number of significant digits. Volume of vinegar (before Trial1 Trial 2 you dilute) USAQ, 5.4 Mass of vinegar (before you 52 dilute) $100 5.20 4.91 Density of vinegar 0.96 0.94 Initial buret level 49.9 MD.IO 0.30 Final buret level 47.90 46.30 Volume of NaOH required for neutralization 47.80 46.00 Final solution color Hot / bright pink Hot / priant pink Show ALL calculations for the following values in the space below Moles of NaOH required for neutralization Moles of acetic acid in sample Mass of acetic acid in sample Molarity of acetic acid (m/m)% of acetic acid in vinegar Calculations 1) Moles of base required for neutralization 2) Moles of acetic acid in vinegar sample (use stoichiometry):

Please answer all with the given data. Thank you

Answer 1

= mass of acetic acid *100 / mass of vinegar sample

Moles of NaOH = 1 Volume of Naor used x Molarity of NaOH solution 44.80 tall x 0.650 M you ) = 0.03107 males Moles of acetic acid = Moles of NaOH consumed . in sample As, according to reaction', CH₂COOH + NaOH CH₃COONa tho one omole q NaOH = one more of che coon. Moles q acetic acid four sample) = 0.03107 moles Mass of acetic acid in sample = Molar mass of x no. of moles acetic acid for acid = 60.09 x B-03107 molées mota = 1.86429 Molarity of acetic acid = Nog mole of acetic and Volume vinegar Solution = 0.03107 moles x 10800 - 5.4 mL 575 M Mass of vinegar - Density of vinegar X Volume of Vinegar. = 6969 X 5.4 mm 5. 184 g (m/m)t of acetic and in vinegar - mik

= 1.864g * 100 / 5,184g

= 35.96%

hope this helps!