# Question f5a94

Jan 17, 2017

Here's what I got.

#### Explanation:

I'll assume that the problem gave you 2.3% as the percent concentration by mass of ethanol in water, meaning that the percent concentration by mass of water is

"% H"_2"O" = 100% - 2.3% = 97.7%

So, a solution's percent concentration by mass, $\text{% m/m}$, sometimes called percent concentration by weight, is calculated by looking at the number of grams of solute present in $\text{100 g}$ of solution.

In your case, a $\text{2.3% m/m}$ ethanol solution contains $\text{2.3 g}$ of ethanol, your solute, for ever $\text{100 g}$ of solution.

Now, molality, $b$, is defined as the number of moles of solute present for every $\text{1 kg}$ of solvent.

If you take a sample of this solution that has a total mass of $\text{100 g}$, you will find that it contains

color(blue)(cancel(color(black)(100))) color(red)(cancel(color(black)("g solution"))) * "2.3 g ethanol"/(100color(red)(cancel(color(black)("g solution")))) = "2.3 g ethanol"

and

color(blue)(cancel(color(black)(100))) color(red)(cancel(color(black)("g solution"))) * ("97.7 g H"_2"O")/(100color(red)(cancel(color(black)("g solution")))) = "97.7 g H"_2"O"

Notice that because the solution contains only ethanol and water, you can find the mass of water by doing

${m}_{\text{water" = "100 g" - "2.3 g" = "97.7 g}}$

Use the molar mass of ethanol to convert the grams to moles

2.3 color(red)(cancel(color(black)("g"))) * "1 mole ethanol"/(46color(red)(cancel(color(black)("g")))) = "0.050 moles ethanol"

Since you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 kg" = 10^3"g}}}}$

you can convert the mass of water to kilograms

97.7 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.977 kg"

This means that the solution's molality will be

b = "0.050 moles ethanol"/"0.977 kg water" = color(darkgreen)(ul(color(black)("0.051 mol kg"^(-1))))#

The answer is rounded to two sig figs.