| Tube | Vol CO(NO3)2 * 6H2O | Vol of Water | Concentration |
| 0 | 0.00 mL | 5.00 mL | ? |
| 1 | 1.00 mL | 4.00 mL | ? |
| 2 | 2.00 mL | 3.00 mL | ? |
| 3 | 3.00 mL | 2.00 mL | ? |
| 4 | 4.00 mL | 1.00 mL | ? |
| 5 | 5.00 mL | 0.00 mL | ? |
The data above is what I did in the lab I need help finding the concentration I know to use the M1V1 = M2V2 equation and the M1 is 0.15 M, V1 is changing, V2 is 50 mL, and M2 is what I need to find which that is the concentration
Solution
M2 =? , V2 = 50 mL
Tube 0
Moles of Co(NO3)2*6H2O = 0 L x 0.15 M =0 moles
total volume= 0mL + 5mL = 5 mL = 0.005L= V1
M1 = 0 moles /0.005L = 0 M
0 M x 5 mL = M2 x 50 mL
M2 = 0 M ( because here no Co(NO3)2*6H2O is present)
Tube 1
Moles of Co(NO3)2*6H2O = 0.001 L x 0.15 M =1.5 x 10-4 moles
total volume= 1mL + 4mL = 5 mL = 0.005L= V1
M1 =1.5 x 10-4 moles /0.005 L = 0.03 M
0.03 M x 5 mL = M2 x 50 mL
M2 = 3 x 10-3 M ( because here 1.5 x 10-4 moles of Co(NO3)2*6H2O is present in 1 mL)
Tube 2
Moles of Co(NO3)2*6H2O = 0.002 L x 0.15 M =3.0 x 10-4 moles
total volume= 2mL + 3mL = 5 mL = 0.005L= V1
M1 =3.0 x 10-4 moles /0.005 L = 0.06 M
0.06 M x 5 mL = M2 x 50 mL
M2 = 6 x 10-3 M ( because here 3.0 x 10-4 moles of Co(NO3)2*6H2O is present in 2 mL)
Tube 3
Moles of Co(NO3)2*6H2O = 0.003 L x 0.15 M =4.5 x 10-4 moles
total volume= 3mL + 2mL = 5 mL = 0.005L= V1
M1 =4.5 x 10-4 moles /0.005 L = 0.09 M
0.09 M x 5 mL = M2 x 50 mL
M2 = 9 x 10-3 M ( because here 4.5 x 10-4 moles of Co(NO3)2*6H2O is present in 3 mL)
Tube 4
Moles of Co(NO3)2*6H2O = 0.004 L x 0.15 M =6.0 x 10-4 moles
total volume= 4mL + 1mL = 5 mL = 0.005L= V1
M1 =6.0 x 10-4 moles /0.005 L = 0.12 M
0.12 M x 5 mL = M2 x 50 mL
M2 = 12 x 10-3 M ( because here 6.0 x 10-4 moles of Co(NO3)2*6H2O is present in 4 mL)
Tube 5
Moles of Co(NO3)2*6H2O = 0.005 L x 0.15 M =7.5 x 10-4 moles
total volume= 5mL + 0mL = 5 mL = 0.005L= V1
M1 =7.5 x 10-4 moles /0.005 L = 0.15 M
0.15 M x 5 mL = M2 x 50 mL
M2 = 15 x 10-3 M ( because here 7.5 x 10-4 moles of Co(NO3)2*6H2O is present in 5 mL)
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If you have any doubts please feel free to ask ...
Tube Vol CO(NO3)2 * 6H2O Vol of Water Concentration 0 0.00 mL 5.00 mL ? 1...