Question

Question #1762a

Answer 1

`"0.80 dm"^3`

Explanation:

The first thing you need to do here is to figure out how many moles of sodium hydroxide are present in `"3.2 g"` of this compound.

To do that, use the compound's molar mass

`3.2 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.080 moles NaOH"`

Now, you know that molarity is defined as the number of moles of solute present in `"1 dm"^3` of solution.

In your case, a `"0.10 mol dm"^(-3)` sodium hydroxide solution contains `0.10` moles of sodium hydroxide, the solute, for every `"1 dm"^3` of solution.

Notice that you need to have `0.080` moles of sodium hydroxide, so you can say that you're definitely going to need less than `"1 dm"^3` of solution.

More specifically, you will need

`0.0080 color(red)(cancel(color(black)("moles NaOH"))) * ("1 dm"^3color(white)(.)"solution")/(0.10color(red)(cancel(color(black)("moles NaOH")))) = color(darkgreen)(ul(color(black)("0.80 dm"^3color(white)(.)"solution")))`

The answer is rounded to two sig figs.