In Analytical Chemistry, we are conducting an Absorbance Lab using CuSO4 . 5H2O . In the lab we are conducting a calibration curve and as part of our prelaboratory procedure we are asked to calculate the molarity of dilutions made from a stock solution; however, I am unsure how the crystalline water affects this calculation or if it is simply a straight forward C1V1=C2V2 calculation. I would really appreciate the help! "Calculate the concentration for each solution that are prepared by the following: Molarity (M) 2.00 mL of stock diluted to 10.00 mL 4.00 mL of stock diluted to 10.00 mL
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"Calculate the concentration for each solution that are prepared by the following:
Molarity (M)
2.00 mL of stock diluted to 10.00 mL
4.00 mL of stock diluted to 10.00 mL
6.00 mL of stock diluted to 10.00 mL
8.00 mL of stock diluted to 10.00 mL
The stock solution is .1000M CuSO4 5H2O.
The molarity of the solution is already given. So we don't need to calculate the molecular weight. we can directly calculate it through V1S1= V2S2 solution.
The stock solution is 1000M CuSO4.5H2O.
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This is somewhat helpful; however, the stock solution is 0.1M, not 1000M and the final answer should be in M (mol/liter) because we are calculating for concentration?
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