How do I arrive at the mass of solvent from a molar fraction concentration?

Question

So the question says:

If the mole fraction for $\ce{H2SO4}$ in an aqueous solution is $0.325$, what is the mass of water (in grams) in $100\ \mathrm{mL}$ of solution?

Me, and my other classmates tried to solve this one but we didn't reach a fully correct answer and our teacher arrives at a different answer.

My answer is:

Since X(H2SO4) + X(H2O) = 1

So: 1-0.325=0.675

X(H2O)=0.675

The mole percent is (0.675*100)=67.5%

1 mole percent = 0.01 mol

H2O moles = 0.675mol

H2O mass (g) = (0.675) * (18) = 12.15 g

Teacher is not convinced with this answer and say its wrong because the total must be 100g, and if the H2O mass is 12.15g then H2SO4 mass will be 31.85g, and 12.25+31.85 dont equal 100.

His answer is simple as follows:

1-0.325=0.675

0.675=67.5% of H2O

H2O mass= 67.5g

Answer 1

Both your teacher and your answer are wrong!

There are two steps to this problem. First, we need to convert between mass fractions and mole fractions. Second, we need to convert from a per-amount basis to a per-volume basis. "Per-amount" is my made up word for all units such as mass fraction, mole fraction, molality, etc that are expressed per amount (whether moles or mass) of substance (whether total or solvent). "Per-volume" units include things like molarity, grams per liter, etc., where the basis is the volume of solution. Converting from an amount basis to a volume basis requires knowing the density of the solution!

Step by step:

1. Converting to mass fraction

$$ 0.325 \frac{\mathrm{mol\;\ce{H2SO4}}}{\mathrm{mol\;total}} \Rightarrow \frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}}$$

$$\frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}} \times \frac{98.1\mathrm{\frac{g\;\ce{H2SO4}}{mol\;\ce{H2SO4}}}}{18\mathrm{\frac{g\;\ce{H2O}}{mol\;\ce{H2O}}}}=2.62\mathrm{\frac{g\;\ce{H2SO4}}{g\;\ce{H2O}}}\Rightarrow \frac{2.62}{2.62+1}\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}=0.724\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}\Rightarrow (1-0.724)\mathrm{\frac{g\;\ce{H2O}}{g\;total}}=0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}$$

This step is pretty easy to do using the information in the problem. There are 72.4 grams of sulfuric acid present in 100 g of the solution. So far, so good.

2. Converting to a per volume basis

But at this point the problem gets very tricky:

...what is the mass of water (in grams) in 100 mL of solution?

It says 100 mL, not 100 g. This makes the problem much harder. This also makes your teacher's answer wrong:

Teacher is not convinced with this answer and says its wrong because the total must be 100g

The total doesn't need to be 100 g because we are apparently dealing with 100 mL of solution. Depending on the density of the solution, the total mass will be more or less than 100 g.

According to Wikipedia, sulfuric acid at a mass fraction of 0.7 has a density of 1.60 kg/L and at a mass fraction of 0.78 has a density of 1.70 kg/L. Let's suppose a mass fraction of 0.724 has a density of ~1.65 kg/L.

$$0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}\times\frac{1650\mathrm{\;g\;total}}{\mathrm{L}}= 46\mathrm{\frac{g\;\ce{H2O}}{L}}$$

With the final number, now it is easy to see that if there are 430 grams of water per liter, then in 100 mL there are 46 grams of water.