Question

R3 was determined to be 24,350 mmHg CO2 per mole of acetic acid using the data collected from the reaction of sodium carbonate with hydrochloric acid in Part A of the experiment. Calculate the concentration in molarity for an acetic acid solution if 105.3 mmHg of pressure was produced from the addition of 6.50 mL of the acid solution to solium carbonate. Select one: a. 1.33 M b. 0.665 M C. 0.0281 M d. 16.2 M e. 0.00432 M

there was no data just this question and ik confused really

Answer 1

P1 = Pressure of CO2/mol acetic acid = 24350 mmHg/mol acetic acid

P2 = Pressure of CO2 produced = 105.3 mmHg

Moles of acetic acid used = P2/P1

= (105.3 mmHg) / (24350 mmHg/mol acetic acid)

n = 0.0043244 mol

Volume = 6.50 mL x 1L/1000 mL

V = 0.0065 L

Molarity of acetic acid solution

= n/V

= (0.0043244 mol) / (0.0065 L)

= 0.665 M

Option B is the correct answer