# Question 608c2

Jan 14, 2018

${\text{0.391 mol kg}}^{- 1}$

#### Explanation:

As you know, a solution's percent concentration by mass tells you the number of grams of solute present for every $\text{100 g}$ of the solution.

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

Consequently, this $\text{100-g}$ 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"))

$\text{mass of solvent = 96.78 g}$

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 $\text{1 kg}$ of solvent.

Since you know how many grams of sodium nitrate are present for every $\text{96.78 g}$ of water, you can calculate the mass of sodium nitrate present for every $\text{1 kg" = 10^3 quad"g}$ 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 ${10}^{3} \quad \text{g" = "1 kg}$ of water, is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 0.391 mol kg}}^{- 1}}}}$

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