What is the molarity of Cu2+ in a catalyzed reaction using 3 drops of CuSO4 at .03M in 50.0ml, and what is the molarity of Cu2+ using 6 drops? (Consider the extent of dilution of the 0.1 M CuSO4 solution in preparing the 50 mL reaction solution. Estimate 20 drops/mL.)
My Work:
3drops*1ml/20drops=.15ml
Total ml=50.15=.05015l
Molarity=moles/L
.03molarity*.050l=.0015molesCuSo4
1moleCu2+ to 1 mole CuSo4 (I think?)
so molarity for 3 drops=.0015/.05015=.03M
I know I did something wrong (I think finding the moles) because I got the same Molarity for both 3 and 6 drops, which I know is wrong. Any help, please and thanks!!!
The problem is worded funny and I can't tell if the final volume is 50 mL or 50.15 mL.
If 50.15 as you are suggesting then
0.03 M x 0.15 mL/50.15 mL = ?M
0.03 M x 0.30 mL/50.30 mL = ?M
If the final volume is 50 mL then
0.03 x 0.15/50 = ?
0.03 x 0.30/50 = ?
To properly calculate the molarity of Cu2+ in the catalyzed reaction using drops of CuSO4, we need to consider the dilution caused by adding drops to a solution.
Let's calculate the number of moles of CuSO4 added when using 3 drops and 6 drops:
3 drops * (1 mL / 20 drops) = 0.15 mL
6 drops * (1 mL / 20 drops) = 0.30 mL
Based on the information given, 20 drops equal 1 mL. So, the volume added can be converted to liters:
0.15 mL = 0.15 mL * (1 L / 1000 mL) = 0.00015 L
0.30 mL = 0.30 mL * (1 L / 1000 mL) = 0.00030 L
Now, let's calculate the molarity using the formula Moles of solute / Volume of solution (in liters):
Moles of CuSO4 added (3 drops) = 0.03 M * 0.00015 L = 0.0000045 moles
Moles of CuSO4 added (6 drops) = 0.03 M * 0.00030 L = 0.0000090 moles
Next, consider that 1 mole of CuSO4 yields 1 mole of Cu2+.
Therefore, the molarity of Cu2+ is equal to the moles of CuSO4 added divided by the total volume of the reaction solution:
Molarity (3 drops) = 0.0000045 moles / (50 mL + 0.15 mL) = 0.0000045 moles / 0.05015 L ≈ 0.0896 M
Molarity (6 drops) = 0.0000090 moles / (50 mL + 0.30 mL) = 0.0000090 moles / 0.05030 L ≈ 0.1786 M
Therefore, the molarity of Cu2+ in the catalyzed reaction is approximately 0.0896 M for 3 drops of CuSO4 and 0.1786 M for 6 drops of CuSO4.