for each solution. Once prepared, you can place Prepare the solutions below usin
ID: 1074388 • Letter: F
Question
for each solution. Once prepared, you can place Prepare the solutions below using a volumetric flask beakers. Be sure to approximately 30 ml of each of these solutions in three separate clean, graduated Solution A: (Together in a 100.00 ml volumetric flask) 0.200 M KIO, 0.50 mL of concentrated H,SO Solution B: (Together in a 100.00 mL volumetric flask) 0.0295 M MnSO H2O 0.150 M Malonic acid solution 3.5 mL 1% stock starch solution (pre-prepared) · . Solution C: Prepare 100.00 ml of a 12% H2O2 (hydrogen peroxide) solution in a volumetric flask from a saturated (stock) 30% H202 solution Lab Report for the Briggs-Rauscher Reaction Name Section Describe below the steps (mass quantities and/or volumes) that you took to create the three solutions for your reaction 1) .Solution A: . Solution B: Solution C:Explanation / Answer
Solution A:
Number of moles of KIO3 in solution = Volume of solution (in L) * Molarity
=> 0.1 * 0.2
=> 0.02 moles
Molar mass of KIO3 = 214 gm/mol
Mass of KIO3 required = number of moles * molar mass = 0.02 mol * 214 gm/mol = 4.28 grams
Volume of concentrated acid required = 0.5 mL
Total volume = 100 mL
100 mL = volume of concentrated acid + Volume of water
Volume of water = 99.5 mL
Hence we can prepare the solution A by adding 4.28 grams of KIO3 in 99.5mL of water and 0.5mL of concentrated H2SO4 solution
Solution B:
number of moles of MnSO4.H2O = volume of solution(L) * molarity(M) = 0.0295 * 0.1 = 0.00295 moles
number of moles of malonic acid = volume of solution(L) * molarity(M) = 0.150 * 0.1 = 0.0150 moles
Mass of MnSO4.H2O = 0.00295 mol * 169 gm/mol = 0.49855 grams
Mass of Malonic acid = 0.0150 mol * 104.0615 gm/mol = 1.56909 grams
Volume of water added = 100 mL - volume of starch solution = 100 - 3.5 = 96.5 mL
so solution is prepared by adding 0.49855 grams of MnSO4.H2O, 1.56909 grams of malonic acid, 3.5 mL of starch solution and 96.5 mL of water
Solution C:
Since the number of moles of H2O2 will remain the same
M1V1 = M2V2
(12/100) * 100 = (30/100) * V2
V2 = 12/30 * 100 = 40 mL
Hence we can prepare the required solution by adding 40mL of 30% stock solution + 60 mL of water
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.