What is the molarity of a sulfuric acid solution prepared by diluting 500.0 mL of 1.00 M H2SO4 to a total volume of 2.50 L?

1 Answer
anor277
Nov 17, 2016

#"Molarity"# #=# #2.00*mol*L^-1# with respect to #"sulfuric acid"#.

Explanation:

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#, and thus has the units of #mol*L^-1#.

So we work out this quotient with respect to the given conditions:

#(500xx10^-3*cancelLxx1.00*mol*L^-1)/(2.50*cancelL)=2.00*mol*L^-1#. At least our answer is dimensionally consistent.

Note that the given concentration is with respect to #"sulfuric acid"#. What would it be with respect to #SO_4^(2-)#, and to #H_3O^+#?

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