E 1. Take three clean test tubes. Label them 1, 2, and 3 2. To test tubes 1 and
ID: 1004547 • Letter: E
Question
E 1. Take three clean test tubes. Label them 1, 2, and 3 2. To test tubes 1 and 3 add about 2 mL. of solution B. To test tube 2 add 2 mL of solution C. 3. To test tube 1 add drops of solution D. To test tubes 2 and 3 add drops of 3M NaOH(aq). In each case add two (2) drops more than just enough to cause a precipitation. Record for each test tube the (total) number of drops added 4. Mix each test tube thoroughly and centrifuge them for about ten seconds. spaced equally. Or Note that the centrifuge must always be balanced. Three test tubes can be two test tubes with the same mass can be opposite one another. To centrifuge a sing balance it by placing a similar test tube with a corresponding amount of water opposite it A reaction is complete when the solution above the precipitate has no more color and/or there is no further precipitation when another drop is added. Cycle through steps 3 and 4 until each le test tube 5. reaction is complete. Record total drops added to complete precipitation. Dispose of the waste carefully as directed. 1. Solution left over from D can go down the drain 2. All solutions and test tubes with iron can go in the "iron waste" bottle. F G Run the computer program to get help with the calculations and attach output to the lab report. III. DATA GI1h-1. volume pomL A. For solution A: 01 volume 1 Mass of graduated cylinder + solution Mass of graduated cylinder + water Mass of graduated cylinder Sttract to get Mass solution = - 51.11g volume (o.0 mL 92 f gradusted eylinder 46.0 Mass solution _-la,so, 12:30 3.0Mass solute _101 -Mass solute ion toget B. For solution B: Volume of solution B made 1002 Volume of solution A used (3 Mass of solution B used for density measurement .5s volume Om Volume of solution Mass of solution B used for density mesurement 55g volume 0omt C. For solution C Volume solution AlamL+ volume solution Boon Which solution is darker? B or Volume solution c il3mL . Drops needed to complete precipitation. Total number of drops added: 3 + 3»13 Total number ofdrops added: 3 + 3+3 t5+6-20 Total number ofdrops added: a + 3 E. Observations: 1. Solution B+ solution D: 2.Solution C + shelfNdH(aq): 3, Solution B+ shelf NaOH (aq):Explanation / Answer
A.(2): As you have calculated, mass of Fe(NO3)3 in the solution = 1.802 g
total mass of the solution = 12.86 g
Hence percentage of Fe(NO3)3 in the solution = (1.802 g / 12.86 g) x 100 = 14.01 %
A.(3): Density of the solution = mass of solution / volume of solution = 12.86 g / 11.2 mL = 1.148 g/mL
A.(4): mass of Fe(NO3)3 in the crystal = 1.802 g
mass of Fe(NO3)3.9H2O crystal = 3.01 g
Hence % of Fe(NO3)3 in Fe(NO3)3.9H2O crystal = (1.802 g / 3.01) x 100 = 59.87%
A.(5): volume of solution = 11.2 mL = 11.2 mL x (1L / 1000 L) = 0.0112 L
moles of solute = 0.00745 mol
Hence molarity of solution = 0.00745 mol / 0.0112 L = 0.6652 M
A.(6): % Fe(NO3)3 in the solution = mass of Fe(NO3)3 / mass of solution)x100 = (1.802 g / 12.86 g) x 100 = 14.01 (w/w)
B: For procedure B:
1: V1 = 8.5 mL, M1 = 0.6652 M
V2 = 100 mL, M2 = ?
M1V1 = M2V2
=> M2 = M1V1 / V2
=> M2 = (0.6652 Mx 8.5 mL) / 100 mL = 0.0565 M
(3): moles of Fe(NO3)3.9H2O in 100 mL of the solution = MxV = 0.0565 M x 0.100 L = 0.0056542 mol
molar mass of Fe(NO3)3.9H2O = 403.99 g/mol
Hence mass of Fe(NO3)3.9H2O need to be added = 0.0056542 mol x 403.99 g/mol = 2.284 g
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