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
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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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.
Well, I screwed up big time. The problem says in big print H2SO4 and I used HCl.
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.
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.
I meant to say that the answer to part 2 isn't the answer to part 1.
To calculate the molarity of a solution, we need to know the mass and the volume of the solute (in moles) as well as the volume of the solution (in liters).
First, we need to find the mass of the solute in the resulting solution. Since equal volumes of the two solutions were mixed, we can assume that the volume of each solution is the same.
Let's assume the volume of each solution is V mL.
Now, let's calculate the mass of H2SO4 in each solution.
For the first solution:
Mass of H2SO4 = 30% by mass × V mL × density = 0.30V × 1.218 g/mL
For the second solution:
Mass of H2SO4 = 70% by mass × V mL × density = 0.70V × 1.610 g/mL
The total mass of H2SO4 in the resulting solution will be the sum of the masses of the individual solutions:
Total mass of H2SO4 = 0.30V × 1.218 g/mL + 0.70V × 1.610 g/mL
Next, we need to determine the volume of the resulting solution. We have been given that the resulting solution has a density of 1.425 g/mL.
Let's assume the volume of the resulting solution is V' mL.
Now we can calculate the mass of the resulting solution:
Mass of resulting solution = V' mL × density = V' mL × 1.425 g/mL
Since the molarity is given by moles of solute divided by liters of solution, we need to convert the masses of the solute and solution into moles.
First, let's calculate the moles of H2SO4 in the resulting solution:
Moles of H2SO4 = (Total mass of H2SO4) / (molar mass of H2SO4)
The molar mass of H2SO4 is:
1 mol of H2SO4 = 2(1.008 g/mol of H) + 32.06 g/mol of S + 4(16.00 g/mol of O) = 98.09 g/mol
Next, we need to calculate the volume of the resulting solution in liters:
Volume of resulting solution = V' mL / 1000 mL/L
Finally, we can calculate the molarity of the resulting solution:
Molarity of the resulting solution = (Moles of H2SO4) / (Volume of resulting solution)
By substituting the values into the formula, we can find the molarity of the resulting solution.