A teacher dilutes 12.0 M HCl down to 2.76 M for a lab. A total of 400.0 ml are needed. How much concentrated acid will be used?

Question

355 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T16:14:33

Under #100*mL#....

Remember it is always ACID TO WATER...

By definition, #"concentration"="moles of solute"/"volume of solution"#..

And thus if the required concentration is #2.76*mol*L^-1#...AND the volume is #400.0*mL#...

#"moles of solute"="concentration"xx"volume"=2.76*mol*L^-1xx400.0*mLxx10^-3*L*mL^-1=1.104*mol#

And so if #12*mol*L^-1# #HCl# is available we need...

#(1.104*mol)/(12*mol*L^-1)xx1000*mL*L^-1-=92.0*mL#...

And please note that the conc. #HCl# we use in the lab is #10.6*mol*L^-1#; this question is NOT chemically sound.