could you please help me with this problem
Thank you.
Prepare 100ml of 2.50M HCl solution for tomorrow's undergraduate experiment. The Stock Solutions cabinet is under the Stockroom Explorer. You will find a 2.50 liter bottle containing 11.6M HCl. Please prepare a flask containing 100ml of a 2.50M solution. To ensure proper credit, please write your calculations in the space given below. Screen capture your solution.
2 answers
You have an 11.6 M HCl solution and you want to use that to prepare 100 mL of 2.50 M HCl solution. You can go about this several ways but here are two of them. The first is to use the dilution formula of mL1 x M1 = mL2 x M2
mL1 x 11.6 = 100 x 2.50
mL1 = 100 x 2.50/11.6 = 21.55.
So you place 21.55 mL of the 11.6 M HCl stock solution into a 100 mL volumetric flask, add distilled or deionized water up to the mark on the volumetric flask, mix thoroughly, stopper and label.
The second is to ask yourself how many moles is in the final solution? That is M x L = 2.50 M x 0.100 L = 0.250 mols.
Then how many L of the 11.6 M HCl stock solution will get 0.250 mols. That is M = mols/L
11.6 = 0.250/L and L = 0.250/11.6 = 0.02155 L oe 21.55 mL.
mL1 x 11.6 = 100 x 2.50
mL1 = 100 x 2.50/11.6 = 21.55.
So you place 21.55 mL of the 11.6 M HCl stock solution into a 100 mL volumetric flask, add distilled or deionized water up to the mark on the volumetric flask, mix thoroughly, stopper and label.
The second is to ask yourself how many moles is in the final solution? That is M x L = 2.50 M x 0.100 L = 0.250 mols.
Then how many L of the 11.6 M HCl stock solution will get 0.250 mols. That is M = mols/L
11.6 = 0.250/L and L = 0.250/11.6 = 0.02155 L oe 21.55 mL.
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