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

1 Answer
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.