Question

Procedure: • Collect about 5 mL of each of the three solutions (IM HCl and 0.005M I/KI) in separate 10 ml beakers. Collect 4M acetone in an Erlenmeyer flask and cover it with a stopper it to avoid evaporation. Perform each of the runs three times. Record the individual time and take the average. Run #1 Use an automatic delivery pipet. Pipet exactly 200 ul acetone, 200 uL HCl and 400ML distilled water into a small test tube. Mix well. When your partner is ready to start, add exactly 200 uL I, and begin the timer. Stir with a small stirring rod to mix thoroughly and remove the rod. The mixture will appear yellow because of 12 color. Place the test tube against a white background (use paper if need be). Record the time in seconds it takes for the yellow color to disappear completely. It will take between 60 and 120 seconds. Record the temperature of the solution after the reaction is over. Dispose of all waste solutions in the appropriate waste bottle. • • Run #2 • Perform run 2 in the same manner as run #1 but introduce 400 L acetone, 200 uL HCL and 200 ML distilled water into a test tube. Mix well. When your partner is ready to start, add exactly 200 L 12 and begin the timer. • Stir with a small stirring rod to mix thoroughly and remove the rod. The mixture will appear yellow because of I, color. Place the test tube against a white background. Record the time in seconds it takes for the

Final M of Acetone, Final M of H+ (hydrogen) and I2 (iodide)?

Answer 1

we know mivi - Movt a here, Mi & Initial molarity Vi - Initial volume M = Final molarity h = Final volume here, & molarity of the species before any reaction tok en place in the réaction mixture. V = 1000 ML , 1000m2 for acetone. Triall 8 = 40 Vis 200m do si vi- s fut - m - m; vi 4 An X 200 ML . Mf = 1 LUTE - 0.8 Torial2 here, initial volume tok en 400 mt. ;. d. 2 x of trial 1 here, final [Acentone ? - 2x 0.8M = 1.6 M. Tria3, 4&5 (: Sarretone was tok en). finas [Acetone) = 0.8m (: Saine inition volume of Tria/3, 4&5

For Ht trial 1 , Vi= 200 ML, Mi=1M = 1000 ML .: M = vi i 2res x 10 = 0.2m torony 0.2M Final [H+] = For trial 12 4 45 inition а) моолле е. 2 мм. So, finas CH) 0.2m for each But for trials the intial volume of Ht was taken 400ML ahich a 2x intial volume of trials for trial 3 final (At) = 2 x 0.2m = 0.400 in for [I] trial 1 Vi= 200 M V = 1000 m Vi x i , M = 0.005 , Mg = ? 200 NT X oors M - . ME V 7ttt = 0.0025M .: final [ = 0.0025 Since for trial 1, 2, 3 &5 the initial volume of u are saine So final [ = d.onses for each But for triat 4 the initial volume of taken & 100 m ie hof initial vol of = 2x 200 = 100 MI So, for trio a Firas [2] = x 0.0025 m = 0.00 125m