# Question #f5a94

##### 1 Answer

Here's what I got.

#### Explanation:

I'll assume that the problem gave you

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

So, a solution's **percent concentration by mass**, *percent concentration by weight*, is calculated by looking at the number of grams of solute present in **of solution**.

In your case, a **for ever**

Now, **molality**, **for every** **of solvent**.

If you take a sample of this solution that has a **total mass** of

#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 containsonlyethanol 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**.