# Question #b22b0

##### 1 Answer

Here's what I got.

#### Explanation:

Start by picking a sample of this solution that contains exactly

#"1 kg" = 10^3color(white)(.)"g"#

of water, the solvent. As you know, the **molality** of a solution tells you the number of moles of solute present **for every**

In your case, a **moles** of sodium hydroxide, the solute, for every

Since we picked a sample that contains **moles** of sodium hydroxide.

Use the **molar mass** of sodium hydroxide to convert the number of mols to *grams*

#2 color(red)(cancel(color(black)("moles NaOH"))) * "40.0 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = "80.0 g"#

This means that the total mass of the solution will be equal to

#overbrace("80.0 g")^(color(blue)("mass of NaOH")) + overbrace(10^3color(white)(.)"g")^(color(blue)("mass of water")) = overbrace("1080 g")^(color(blue)("mass of solution"))#

Now, in order to find the solution's **mass by mass percent concentration**, **of the solution**.

To do that, use the known composition of the sample

#100 color(red)(cancel(color(black)("g solution"))) * "80.0 g NaOH"/(1080color(red)(cancel(color(black)("g solution")))) = "7.41 g NaOH"#

Since this represents the mass of sodium hydroxide present in

#color(darkgreen)(ul(color(black)("% m/m = 7.4% NaOH")))#

I'll leave the answer rounded to two **sig figs**, but do not forget that you have one significant figure for the molality of the solution.

So the answer *should* be given as

#"% m/m = 7% NaOH"#