DATA Be sure to use appropriate significant figures! A. For Solution A Volume of
ID: 945610 • Letter: D
Question
DATA Be sure to use appropriate significant figures! A. For Solution A Volume of Solution A prepared: 100.0 mL Mass of Fe(NO), 9H20 solid: For Solution B: 1. B. 30.80 Volume of Solution B prepared: 200.0 mL Volume of Solution A used: Density measurement Mass of Solution B used: 1 2. 2.OD Volume of Solution B used:20.L Volume of Solution B used: C. For Solution C: Volume of Solution A used Volurme of Solution B used: Volume of Solution C prepared: -05_ 252 8-5 me (from the pipette reading) (from the pipette reading) (from the pipette reading) (from the graduated cylinder reading) D. Absorbance measurements: 1. Observations-describe the relative depth of color of the solutions in the cuvettes: in Coa 2. Spectrophotometer Data Absorbance at 480nm Solution A Solution B Solution C 210 0114 O. 628 Remember: work alone and show your work DATA ANALYSIS A. For Solution A: 1. Use the mass of the hydrate you weighed (404.0 g/mol) to calculate the of moles of Fe(NOs)s in the solid hydrate 2. 2. Calculate the molarity of Fe(NOaja in Solution A 3. How many mL of your Solution A would be needed to react with 150.0 mL. of 0.400 M NaOH? Use dimensional analysis. 4. Calculate the molarity of nitrate ion in Solution AExplanation / Answer
3) how many moles of solution A required to react completely with 150.0 ml of 0.400 M naoh
= 150.0milli * 0.400 M = 60 milli moles
concertration of soution A = 0.0374 M * volume of solution A = 60 milli moles
or volume of solution A = 60 / 0.0374 = 1604.27 ml
4) molarity of nitrate
as Fe(NO3)2.9H2O gives ----> 3 moles of nitrate ions
molarity of NO3- = 3 * molarity of Fe(NO3)3.9H2O
= 3* 0.0374 = 0.1122 M
5) % of feNO3
= mass of feNO3 / volume of solutio n * 100 = (1.51 / 100 )* 100 = 1.51 %
B ) a) M1V1 = M2V2
molarity of solution A* volume of A molarity of B * volume of B
0.0374 * 0.0308 Lit = MB * 0.2 LIt
molarity of B = 0.0057 M
b)density = mass/ volume = 18.12 / 20 ml = 0.906 g/ml
as density * volume = mass of solute= 0.00516
C) (w/w%) = given mass / mass of solution * 100 = 0.00516 / 0.2 *100= 2.58 %
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