# Question #65301

##### 1 Answer

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

So, if you have

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

Now, **molarity** is defined as the number of moles of solute present in

To make the calculations easier, let's assume that you have a sample of

This sample is **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

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

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