# Question 1762a

Mar 13, 2017

${\text{0.80 dm}}^{3}$

#### Explanation:

The first thing you need to do here is to figure out how many moles of sodium hydroxide are present in $\text{3.2 g}$ 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 ${\text{1 dm}}^{3}$ of solution.

In your case, a ${\text{0.10 mol dm}}^{- 3}$ sodium hydroxide solution contains $0.10$ moles of sodium hydroxide, the solute, for every ${\text{1 dm}}^{3}$ 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 ${\text{1 dm}}^{3}$ of solution.

More specifically, you will need

$0.0080 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaOH"))) * ("1 dm"^3color(white)(.)"solution")/(0.10color(red)(cancel(color(black)("moles NaOH")))) = color(darkgreen)(ul(color(black)("0.80 dm"^3color(white)(.)"solution}}}}$

The answer is rounded to two sig figs.