Question

Molarity Equation.pdf - Google Drive Gram Equation.pdf - Google Drive Chemistry Labi - 7... ignment Exp. 12 Time Left:0:51:25 Madalyn Stewart: Attempt 2 Question 5 (1 point) The average molarity of acetic acid (HC2H302) in vinegar is determined as 0.845 M. If 5.00 mL of vinegar was used in the titration, how many grams of acetic acid were present in vinegar? If you need help, click on the link: Grams from molarity and volume 0.169 g O 1.69 x 10-4 0.2548 2548 4.25

Answer 1

0.845 m acetic acid solution , in vinegar means - 1000 mL vinegar contains 0, 845 mole acetic acid i 5 mL " " 0.845 xE) more aeltic and 1000 = 4.225x10 mole acetic acid Molar mass of acetic acid = 60.05 glmol. : mass of acetic and present = no. of moles - mass present x acetic molar acid of = 4.225 x 10 3 x 60.05 g = 2.537x10 g ² 0.254 g so option-e is the answer.