please show me how you calculate the table the experiment is Cyclic Volumetric (
ID: 1024307 • Letter: P
Question
please show me how you calculate the table
the experiment is Cyclic Volumetric
(10 pts) Provide the weight or volume missing in the Table below in preparation for the experiment
#
Reagent
Volume
Molarity
Weight or Volume
Solvent
1
K3Fe(CN)6
100 mL
50mM
1mM KNO3
2
KNO3
500 mL
0.1 M
H2O
3
KCl
500 mL
0.1 M
H2O
4
K3Fe(CN)6
50 mL
2 mM
0.1 M KNO3
5
K3Fe(CN)6
50 mL
4 mM
0.1 M KNO3
6
K3Fe(CN)6
50 mL
6 mM
0.1 M KNO3
7
K3Fe(CN)6
50 mL
8 mM
0.1 M KNO3
8
K3Fe(CN)6
50 mL
10 mM
0.1 M KNO3
9
K3Fe(CN)6
50 mL
2 mM
0.1 M KCl
10
K3Fe(CN)6
50 mL
4 mM
0.1 M KCl
11
K3Fe(CN)6
50 mL
6 mM
0.1 M KCl
12
K3Fe(CN)6
50 mL
8 mM
0.1 M KCl
13
K3Fe(CN)6
50 mL
10 mM
0.1 M KCl
#
Reagent
Volume
Molarity
Weight or Volume
Solvent
1
K3Fe(CN)6
100 mL
50mM
1mM KNO3
2
KNO3
500 mL
0.1 M
H2O
3
KCl
500 mL
0.1 M
H2O
4
K3Fe(CN)6
50 mL
2 mM
0.1 M KNO3
5
K3Fe(CN)6
50 mL
4 mM
0.1 M KNO3
6
K3Fe(CN)6
50 mL
6 mM
0.1 M KNO3
7
K3Fe(CN)6
50 mL
8 mM
0.1 M KNO3
8
K3Fe(CN)6
50 mL
10 mM
0.1 M KNO3
9
K3Fe(CN)6
50 mL
2 mM
0.1 M KCl
10
K3Fe(CN)6
50 mL
4 mM
0.1 M KCl
11
K3Fe(CN)6
50 mL
6 mM
0.1 M KCl
12
K3Fe(CN)6
50 mL
8 mM
0.1 M KCl
13
K3Fe(CN)6
50 mL
10 mM
0.1 M KCl
Explanation / Answer
Solved first 6 parts, please post multiple question to get the remaining answers
1) Molar mass of K3Fe(CN)6 = 329.24 g/mol
Number of moles of K3Fe(CN)6 = 100/1000 * 50mM = 5mM = 5 * 10^(-3) moles
mass of K3Fe(CN)6 = 1.6462 grams
2) Molar mass of KNO3 = 101.1032 g/mol
Number of moles of KNO3 = 500/1000 * 0.1M = 0.05 moles
mass of KNO3 = 0.05 mol * 101.1032 gm/mol = 5.05516 grams
3) Molar mass of KCl = 74.5513 g/mol
Number of moles of KCl = 500/1000 * 0.1 = 0.05 moles
mass of KCl = 0.05 mol * 74.5513 gm/mol = 3.727 grams
4) Molar mass of K3Fe(CN)6 = 329.24 g/mol
Number of moles of K3Fe(CN)6 = 50/1000 * 2mM = 10^(-4) moles
mass of K3Fe(CN)6 = 0.032924 grams
5) Molar mass of K3Fe(CN)6 = 329.24 g/mol
Number of moles of K3Fe(CN)6 = 50/1000 * 4mM = 2 * 10^(-4) moles
mass of K3Fe(CN)6 = 0.065848 grams
6) Molar mass of K3Fe(CN)6 = 329.24 g/mol
Number of moles of K3Fe(CN)6 = 50/1000 * 6mM = 3* 10^(-4) moles
mass of K3Fe(CN)6 = 0.098772 grams
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