Question

B. Analysis of Vinegar (Acetie Acid) Trial aceti ic aci S Advertised % Acetic Acid Trial #2 Trial #3 NAT of vi 8. Volume of vinegar 10 ml 10ml 1.0 ml 9. Final buret reading 6.4 _ml 10. Initial buret reading 0.00 mL 6.4 ml. 13.4_mL 6.4 ml 7.0 ml 20.2__ml 13.4 ml 6.8 mi 11. Volume of NaOH 12. Molarity of vinegar (Show set-up) - M - 13. Average molarity of vinegar 14. Percent acetic acid in vinegar (Show set-up)

Answer 1

Acetic And H₂ soy (sulfuric acid) Trial 1 Molarity NaOH = 0.145 M Volume Nath = 6.4mL Volume Vinegar = 1 mL CH₂COOH + NaOH → CH₂ Coona + H2O Mix = M2 VL Vinegar Nauh Mvin. = 0.145M *6.400 Inh Muina 0.928 M) Vaud lomL Vol NaOH = 14.5mL Molarity Nath = 0.145M H₂ soy taNach Nasoy +2420 0:2) 2x Mix = M2V₂ x Aud Base 2x Marid X loml = 0.145 x 14.57L I Mauid = 0.105125M! Trial 2 Similarly, Trial 2 Similarly, Muin = 0.145M* 7 ml 1mz Mrin = 1.015M] Mauid = 0.145 M 13.6mL 2x 10mL Macid = 0.0986M ! Trial3 Trial 3 Mvin. = 0.145m x 6.8ML 1 ml Mauid = 0.145M X 14:1 ml 2x lomx Mauid = 0.102225M! Muin. = 0.986 MI Average Molarity - Trial 1+2+3 Average Modernity - Triall +2+3 = 0.105125 +0. 0906+0.02225 3 = 0.928+1:05+0.986 [ Marg. = 0.101017 Thur, average nolarity of acid was found to be 0.101017 | Mavg.= 0.9763M 1 looomL of solution contains 0.9763 md. Acetic and loomL " "0.09763 mol " mass= moles x molar mass = 0.09763 hal x 60 g = 5.856 g loomL vinegar Contains 5.856 g Acid 17:= 5,856. auid. I Massy