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Mole fraction of ethanol in a solution with water is $0.040$.

What will be the molarity of ethanol in water? Density of water can be assumed to be $1\ \mathrm{g/mL}$.

This is how I tackled the problem:

$0.040 = n(\mathrm E)/(n(\mathrm E) + n(\mathrm W))$

where $n(\mathrm E)$ is number of moles of ethanol and $n(\mathrm W)$ is number of moles of water. This is the first equation.

I assumed that there is $1\ \mathrm L$ of solution in total and that the molarity is to be found at STP.

That gives:

$(n(\mathrm E) + n(\mathrm W)) \times 22.7 = 1$

This is the second equation.

Two equations, two unknowns. Solving yields, $n(\mathrm E) = 0.00176$ and because the total volume of the solution is $1\ \mathrm L$, molarity is $0.00176\ \mathrm M$.

But that answer is wrong! Probably because I am assuming molarity is to be found at STP.

Here is an answer I found online:

Assume there is $1\ \mathrm L$ water. This implies $1000\ \mathrm g$ or $55.56\ \mathrm{mol}$ water.

Using $\text{eq(1)}$ from my answer, $n(\mathrm E) = 2.32$.

And then the answer concluded by saying that as there is $1\ \mathrm L$ water, molarity is the same, $2.32\ \mathrm M$.

How can this be? Molarity is moles over total volume not just the volume of the solvent.

Am I missing something or the online answer is wrong? Is my answer correct? If not, what's the correct solution?

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  • $\begingroup$ Curious, where did the 22.7 come from? $\endgroup$ – MaxW Sep 1 '18 at 15:44
  • $\begingroup$ @MaxW The volume of 1 mole of an ideal gas at 273 K and 1 bar. $\endgroup$ – user64082 Sep 1 '18 at 16:19
  • $\begingroup$ @user64082 You might also want to check your units. Especially your values fore the amount of substance $n$ sometimes are correctly expressed in mol, but also wrongly in M = mol/l or without any unit. Besides, the quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’. $\endgroup$ – Loong Sep 1 '18 at 16:24
  • $\begingroup$ @Loong Thanks for correcting me. I have edited my question so that n(x) is strictly a number without units. $\endgroup$ – user64082 Sep 1 '18 at 16:32
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    $\begingroup$ @user64082 - the solution is a liquid. The molar volume of a gas has nothing to do with this problem. $\endgroup$ – MaxW Sep 1 '18 at 16:36
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For a mass of 1000 grams of solution we can find the number of moles of water and ethanol. Let $x$ be the total number of moles of ethanol and water in 1000 grams of solution. Then

$$(0.040x)46.07 + (0.960x)18.015 = 1000$$ $$x = 52.254$$

So there are $0.040 \times 52.254 = 2.09$ moles of ethanol in the 1000 grams of solution, and $0.96 \times 52.254 = 50.16$ moles of water.

However there is no way to compute molarity since the density of the ethanol-water solution itself is unknown.

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  • $\begingroup$ The density of the solution is not 1 g/mL. $\endgroup$ – user64082 Sep 1 '18 at 16:18
  • $\begingroup$ Ahhhh... misread the problem statement. $\endgroup$ – MaxW Sep 1 '18 at 16:21
  • $\begingroup$ No problem. Happens to everybody :) $\endgroup$ – user64082 Sep 1 '18 at 16:23
  • $\begingroup$ @user64082 - I edited my answer. The problem is unsolvable as given. The only way to solve it would be to look up the density of ethanol-water mixtures. Are you sure that the problem didn't ask for molality rather than molarity? $\endgroup$ – MaxW Sep 1 '18 at 16:29
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    $\begingroup$ Again, the problem is unsolvable as written, unless you look up the density of the particular ethanol-water mixture. $\endgroup$ – MaxW Sep 1 '18 at 16:45

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