What would be the molarity of solution obtained by mixing equal volumes of 30% by mass H2SO4 (d=1.218 g/mL) and 70% by mass H2SO4 (d=1.610 g/mL). If the resulting solution has density 1.425 g/mL, calculate its molarity

Answer 1

Equal volumes will average the % HCl so (30+70)/2 = 50% HCl.
mols in 1000 mL will be
1.425 g/mL x 1000 mL x 0.50 x (1/36.5) = ?
And that's the M since mols/L = M.

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Answer 2

The first question asks for the MOLARITY of the solution after adding the two together; the second question asks for the molarity using the DENSITY. You will get two different answers. Also, the acid that is given is H2SO4 and not HCl. The molecular weight that Drbob222 used to solve the problem will return the incorrect answer. Also, the addition of the two together will result in the acid being slightly higher than 50%, based on the density of the two acids.

Answer 3

Well, I screwed up big time. The problem says in big print H2SO4 and I used HCl.

Answer 4

part 2.
I looked up the %H2SO4 in density of 1.425 g/mL and it is 52.63%.
M H2SO4, then, is
1.425 g/mL x 1000 mL x 0.5263 x (1/98) = mols in 1 L and that is M.

Answer 5

I agree with DrBob222 and then I don't agree with him: I agree that the 2nd part that he gave you in his last point is correct, but the answer to part 2 isn't the answer to part 2. I agree that it should because adding these two concentrations of acid together doesn't produce a TOTAL VOLUME that is additive. For example, if you were to add 100mL of both acids together, the total volume should not be 200mL. But the way that the problem is worded, this is what I think that author of the question wants you to do for the first part. However, this is a poorly worded question and the second part is the correct way to determine the molarity of the acid, and not the way that it is implied in the first part.

Answer 6

I meant to say that the answer to part 2 isn't the answer to part 1.