# Question 7cad8

Dec 26, 2015

${\text{0.10 mol dm}}^{- 3}$

#### Explanation:

Your strategy here will be to use the molar mass of the compound and the sample given to you to find how many moles of sodium nitrate, ${\text{NaNO}}_{3}$, you will have in solution.

Once you know how many moles of you have, use the volume of the solution to calculate its molarity.

So, a substance's molar mass tells you the mass of one mole of that substance. In your case, sodium nitrate is said to have a molar mass of $\text{85 g/mol}$, which means that every mole of this compound has a mass of $\text{85 g}$.

This means that your $\text{1.7-g}$ sample will contain

1.7 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_3/(85 color(red)(cancel(color(black)("g")))) = "0.020 moles NaNO"_3#

Now, it's important to realize that

$\text{1 dm"^3 = "1 L}$

This means that your solution will have a volume of $\text{0.20 L}$.

As you know, molarity is defined as moles of solute, which in your case is sodium nitrate, divided by liters of solution.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

Since you have everything that you need, plug your values into this equation and find the molarity of the solution

$\textcolor{b l u e}{c = \frac{n}{V}}$

$c = \text{0.020 moles"/"0.20 L" = "0.10 mol L"^(-1) = "0.10 mol dm"^(-3) = "0.10 M}$

SIDE NOTE As a fun fact, your solution will not actually be
$\text{0.10 M}$ sodium nitrate because sodium nitrate dissociates completely in aqueous solution to form

${\text{NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

Since you have $1 : 1$ mole ratios between all three chemical species, you can say that your solution will be $\text{0.10 M}$ in ${\text{Na}}^{+}$ and $\text{0.10 M}$ in ${\text{NO}}_{3}^{-}$.

If you want, you can read more on this concept here

https://socratic.org/questions/what-is-formality-explain-with-an-example