Question

Question 1 (10 points) If 10.46 mL of an acetic Acid solution required 9.17 mL of 0.9479 M NaOH to reach the endpoint, what's the concentration of the Acetic Acid solution? Your Answer: Answer Question 2 (10 points) If 9.26 mL of an acetic Acid solution required 9.90 mL of 1.0739 M NaOH to reach the endpoint, what's the concentration of the Acetic Acid solution? Your Answer: Answer Question 3 (10 points) 5% vinegar is approximately -____molar.

Answer 1

using the M1V1=M2V2 to determine all.

Acetic acid M, -4 V = 10.46mL Naon M₂ = 0.9479 M. V = 9-17 mL Miv, = M₂ V₂ M = M₂ V₂ = 0.9479m x g. 17mL 10.46mL - 10.831_n Accliclaced Naon My = 1.07399 V, -9.9 mL V-9.26 m2 Mr-Mar M. = M = M2 V ₂ M2 2 = 1.0739x9.9 = 9.26 Mi = 11.14817

[ molar mass of Ch₂coon=60.052g/mol in 100g to = 100mL. Date 57. Cacoon - means 59 Chacoon - density of H₂O = 1g/mL. volume of W, 0 100mL. Tholes of Cly coou = so 20.08326 phobiny Lawyeon - 0100326 rolex recome-to-83267 1000mL 0.08326 100ml 32 12