QUESTIONS 1. Vinegar contains acetic Acid, CH3COOH. Titration of 5.00g of vinegar with 0.100 M
NaOH requires 33.0cmº to reach the equivalence point.
(a) What is the weight percentage of CH3COOH in vinegar?
(b) If the vinegar has a density of 1.005gcm", what is the molarity of CH3COOH in
vinegar?​

389929.827

Explanation:

33806.78° C

(a) 3.96 (w/w) %

(b) 0.663M

Explanation:

The acetic acid, CH₃COOH, reacts with NaOH as follows:

CH₃COOH + NaOH → CH₃COONa + H₂O

Where 1 mole of acetic acid reacts with 1 mole of NaOH

In the equivalence point of the titration, the moles of acetic acid are equal to moles of NaOH added. To solve this question we need to find the moles of NaOH and thus, the mass of acetic acid

Moles NaOH = Moles acetic acid:

33.0cm³ = 0.033L * (0.100moles / L) = 0.0033 moles

Mass acetic acid -Molar mass: 60g/mol-:

0.0033 moles * (60g / mol) = 0.198g

a) Weight percent is 100 times the ratio of the mass of acetic acid (0.198g) in the mass of vinegar (5.00g):

0.198g / 5.00g * 100 = 3.96 (w/w) %

(b) The volume of the vinegar in the 5.00g is:

5.00g * (1mL / 1.005g) * (1L / 1000mL) = 0.00498L

And the molarity is:

0.0033moles / 0.00498L = 0.663M

43.452

step-by-step explanation:

284 times .153 = 43.452

am i too late?

explanation: