Question

answers will vary for this question.) (a) Using the average molarity value you reported for your unknown acetic acid solution in Part 2, how many grams of acetic acid would be present in 1.00 L of your unknown acetic acid solution? 0.295 mol HCH₂O2/1oox 60.069 HC2H₂O I T Imoltte H₂O2 = 17,79 HC₂H₂O₂ L 17.79 HGHz Oz (b) Convert the molarity of your unknown acetic acid solution into a mass percent of acetic acid. Assume that the density of your unknown acetic acid solution is 1.01 g/mL molarity 0.295m acetic acid 2. A 30.0 mL sample of 0.220 M sulfuric acid is titrated with a potassium hydroxide solution. If it takes 26.4 mL of base solution to reach the endpoint, what is the molarity of the base solution?

Answer 1

Modari knoon acetic acide 0.295M density - 1.0 g/mL Mw (Hey HSO ) = 60.05 g/mol Se og Hells rool (* So 0.295 m Hesto = 1.00L ( 0.295 ms ) (60.052) - 20.05 moi IL Tomor = 17.719 of acetic acid and in solution - 1.004(100ml). (1002) mous % 2 1010 g f acetic acid - [17.71 Llorot 17.71) x100 % = 107232 2) Let us consider numbers of moles of base presents 30.0 x 0.20 = 26.4 xs + HO + 3 5 30.08 0.2 82 32 30-0x0.2 - 0.227 moles 20.227 moles 26.4 So molarity of base a 1.0.227 (0.03 +0.0264) 4.024 mol/lit 4,024 M

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