A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

1 Answer
anor277
May 20, 2017

We use the relationship, "concentration"="moles of solute"/"volume of solution" TWICE to get a volume of ..............0.25*L

Explanation:

"concentration"="moles of solute"/"volume of solution"

And thus "moles of solute"=6*mol*L^-1xx0.500*L=3*mol required initially.

Since this comes from a volume of 12*mol*L^-1 HCl.....we get

(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L

The concentrated hydrochloric acid used in the lab ("36%w/w") is 10.6*mol*L^-1 so this is another poor question.

Related questions
See all questions in Solution Formation
Impact of this question
8362 views around the world
You can reuse this answer
Creative Commons License