Making a 10L HCl solution with 8.77% HCl in a 100 mL solution from a 16 M stock solution

Question

How would you make 10L of HCl solution with 8.77% HCl in 100 mL from a 16M stock solution.

Here's what I have tried so far:

I put my percent HCl and got that down to match with the moles of HCl I had. Then I took 1.754% of 20 ml and converted that into liters. Then I found the molarity of the HCl I need to turn the 16 M solution into. Finally I did a dilution calculation to find the volume of water needed to make the 16 M stock solution into the starting percentage of HCl.

I am fairly certain that I am diluting with water, I don't know what else it would be with.

Is this the right way to go about this, or do I need to restart. Any help is much appreciated, and thanked for in advance!

Answer 1

I figured out the problem I had. I needed a way to convert grams to milliliters, which was density.

I found the density of HCl for the 8.77% concentration: $$ HCl \ Density = 1.04 g/mL$$ I then found how much HCl would be in 100g of solution: $$ 8.77\% \ of \ 100g, \ so \ 8.77g \ HCl \ and \ 91.23g \ solvent$$ I did some dimensional analysis to get to moles divided by liters, so that I could get the molarity: $$ (\frac{8.77 g \ HCl}{1})*(\frac{1 \ mol}{36.461g})*(\frac{1}{100g \ solution})*(\frac{1.04g}{1 mL})*(\frac{1000 mL}{1 L}) = 2.5 M$$ Then, I did a nice dilution calculation: $$ 10 L *2.5M = 16M *L_2$$ $$ \frac{25}{16} = 1.5625 L$$

To do the dilution I would place 1.5625 L HCl into a beaker, and then fill with solvent until I reached a total solution volume of 10 L.