Question

Question #65301

Answer 1

Here's what I got.

Explanation:

For starters, I'll assume that your ethanol solution is labeled using US standards, which have alcohol proof defined as twice the value of alcohol by volume, ABV.

Alcohol by volume is essentially the volume by volume percent concentration of the solution, i.e. the number of milliliters of ethanol, the solute, present in `"100 mL"` of solution at room temperature.

So, if you have `"100-proof"` alcohol, you can say that

`"100-proof" = 2 * "ABV" implies "ABV" = 50%`

Now, molarity is defined as the number of moles of solute present in `"1 L" = 10^3` `"mL"` of solution.

To make the calculations easier, let's assume that you have a sample of `10^3` `"mL"` of `"100-proof"` alcohol.

This sample is `50%` by volume ethanol, so you can say that it contains

`10^3 color(red)(cancel(color(black)("mL solution"))) * overbrace("50 mL ethanol"/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)(="50% ABV")) = "500 mL ethanol"`

Use the density of the ethanol to figure out the number of grams of solute present in this sample

`500 color(red)(cancel(color(black)("mL ethanol"))) * overbrace("0.789 g"/(1color(red)(cancel(color(black)("mL ethanol")))))^(color(blue)(" = 0.789 g mL"^(-1))) = "394.5 g"`

Finally, to convert this to moles, use the molar mass of ethanol

`394.5 color(red)(cancel(color(black)("g"))) * "1 mole ethanol"/(46.07color(red)(cancel(color(black)("g")))) = "8.56 moles ethanol"`

Since this represents the number of moles of ethanol present in `10^3` `"mL"` of `"100-proof"` alcohol, you can say that the molarity of the solution is equal to

`color(darkgreen)(ul(color(black)("molarity = 8.6 mol L"^(-1))))`

I'll leave the answer rounded to two sig figs.