Calculating Volume of 6.0M H2SO4 to Make 0.25M Solution

[scavok]

I've been trying to figure this problem out for atleast an hour, I just don't understand how to get the volume. I can't find any examples of such problems in my textbooks, only problems to get the mass, not volume.

Here's the question verbatim:

What volume (mL) of 6.0M H₂SO₄ should be taken in order to prepare 50.0 mL of a solution which is 0.25M in H₂SO₄?

The molecular weight of H₂SO₄ is 98.078amu (g/mol)

I don't really even know where to go from there. I guess you would need the amount of grams in .25M of H₂SO₄, which is 24.520g. But how the heck do you get volume? You would need the density, wouldn't you?

Maybe if someone could just rewrite the question using different terms I would understand it. It doesn't even seem like it's written in english to me. Particularly the "which is 0.25M in H₂SO₄" part.

Thanks for the help

 

[cepheid]

It looks like you're having a problem with terminology here. The problem does give you all of the information you need to solve it. It can be easily translated into english ( ) from the fairly commonplace chemistry shorthand used to indicate solution concentrations. 6.0M H₂SO₄ should be read as "six molar sulfuric acid". What is this "molar"? It is a unit for concentration, a shorthand for moles per litre! Now the problem should make sense, as you have been given the initial and desired final solution concentrations.

 

[thunderfvck]

c1v1 = c2v2

c1: initial concentration
c2: final concentration
v1: initial volume
v2: final volume

Keep your units consistent.

 

[chem_tr]

As an addition to the friends posted, molarity times volume gives millimoles if M and mL are used, respectively. You'll get moles if M and L are used (resp.). I think it would be better to quote this:

$$molarity=\frac{moles}{liters} or \frac{millimoles}{milliliters}$$