How do we prepare #2*mol*L^-1# #HCl# from #16*mol*L^-1# #HCl#?

1 Answer
anor277
Sep 2, 2017

We use....approx. #1/8*L# of the stuff........

Explanation:

We use the definition of concentration......

#"Concentration"="Moles of solute"/"Volume of solution"#

Now we want #500*mLxx10^-3*L*mL^-1xx4*mol*L^-1#. Agreed?

And thus here the moles of solute ...........#500*cancel(mL)xx10^-3*L*cancel(mL^-1)xx4*mol*cancel(L^-1)=2*mol#.

We have #16*mol*L^-1# #HCl# available.....and so from the equation above....

#"Volume"="Mole of solute"/"Concentration"=(2*mol)/(16*mol*L^-1)=1/8*L#

#=0.125*L#

Just to add, that this question is chemically unrealistic. The strongest hydrochloric acid you can buy commercially is approx. #"34%(w/w)"#; this has a molar concentration of #12*mol*L^-1#. And practically, when you mix acid and water, you ALWAYS add acid to water.

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