# Question 5a849

Mar 3, 2017

${\text{0.1 mol kg}}^{- 1}$

#### Explanation:

We use molality as a measure of the number of moles of solute present in a solution for every $\text{1 kg}$ of solvent.

In your case, the molality of the solution will tell you how many moles of sodium hydroxide, your solute, you'd get for every $\text{1 kg}$ of water, your solvent.

You already know that a sample of this solution contains $\text{2 g}$ of sodium hydroxide in $\text{500 g}$ of water, so the first thing to do here is to figure out how much sodium hydroxide you'd get for $\text{1 kg}$ of water.

1 color(red)(cancel(color(black)("kg water"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "2 g NaOH"/(500color(red)(cancel(color(black)("g water")))) = "4 g NaOH"

Now all you have to do is convert the mass of sodium hydroxide to moles by using the compound's molar mass

4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.1 moles NaOH"#

Since this represents the number of moles of solute present in $\text{1 kg}$ of water, the molality of the solution will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 0.1 mol kg}}^{- 1}}}}$

The answer is rounded to one significant figure.