Question #1762a

1 Answer
Stefan V.
Mar 13, 2017

#"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.

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