Question

Question #5a849

Answer 1

`"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 `"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 `"1 kg"` of water, your solvent.

You already know that a sample of this solution contains `"2 g"` of sodium hydroxide in `"500 g"` of water, so the first thing to do here is to figure out how much sodium hydroxide you'd get for `"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 `"1 kg"` of water, the molality of the solution will be

`color(darkgreen)(ul(color(black)("molality = 0.1 mol kg"^(-1))))`

The answer is rounded to one significant figure.