Question

Question #f5a94

Answer 1

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, `"% m/m"`, sometimes called percent concentration by weight, is calculated by looking at the number of grams of solute present in `"100 g"` of solution.

In your case, a `"2.3% m/m"` ethanol solution contains `"2.3 g"` of ethanol, your solute, for ever `"100 g"` of solution.

Now, molality, `b`, is defined as the number of moles of solute present for every `"1 kg"` of solvent.

If you take a sample of this solution that has a total mass of `"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_"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

`color(blue)(ul(color(black)("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.