Question

Question #608c2

Answer 1

`"0.391 mol kg"^(-1)`

Explanation:

As you know, a solution's percent concentration by mass tells you the number of grams of solute present for every `"100 g"` of the solution.

In your case, the solution is said to be `3.22%` sodium nitrate by mass, which means that if you take `"100 g"` of this solution, the sample will contain `"3.22 g"` of sodium nitrate, the solute.

Consequently, this `"100-g"` sample of the solution will contain

`"mass of solvent" = overbrace("100 g")^(color(blue)("mass of solution")) - overbrace("3.22 g")^(color(blue)("mass of solute"))`

`"mass of solvent = 96.78 g"`

of water, the solvent. Now, in order to find the molality of the solution, you need to figure out how many moles of sodium nitrate are present for every `"1 kg"` of solvent.

Since you know how many grams of sodium nitrate are present for every `"96.78 g"` of water, you can calculate the mass of sodium nitrate present for every `"1 kg" = 10^3 quad"g"` of water first

`10^3 color(red)(cancel(color(black)("g water"))) * "3.22 g NaNO"_3/(96.78color(red)(cancel(color(black)("g water")))) = "33.27 g"`

then use the molar mass of sodium nitrate to convert the number of grams to moles.

`33.27 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_3/(84.9947color(red)(cancel(color(black)("g")))) = "0.391 moles NaNO"_3`

You can thus say that the molality of the solution, i.e. the number of moles of sodium nitrate present for every `10^3 quad "g" = "1 kg"` of water, is equal to

`color(darkgreen)(ul(color(black)("molality = 0.391 mol kg"^(-1))))`

The answer is rounded to three sig figs, the number of sig figs you have for the percent concentration of the solution.