# Question 324ec

Mar 14, 2017

${\text{260 g NaNO}}_{3}$

#### Explanation:

The thing to remember about a solution's molarity is that it's supposed to tell you the number of moles of solute present in exactly $\text{1 L}$ of solution.

In this case, a $\text{1.5-M}$ sodium nitrate solution will contain $1.5$ moles of sodium nitrate, the solute, for every $\text{1 L}$ of solution.

In other words, regardless if you scale this solution up or down, i.e. if you make more than $\text{1 L}$ of less than $\text{1 L}$, it must always contain $1.5$ moles of solute per liter.

This implies that you can use molarity as a conversion factor to go from the volume of the solution to the number of moles it contains or vice versa.

In your case, the solution has a total volume of $\text{2 L}$, so it must contain twice as many moles as $\text{1 L}$ of the same solution

2.0 color(red)(cancel(color(black)("L solution"))) * overbrace("1.5 moles NaNO"_3/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(="1.5 M NaNO"_3)) = "3.0 moles NaNO"_3#

To convert the number of moles to grams, use the compound's molar mass

$3.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles NaNO"_3))) * "85.0 g"/(1color(red)(cancel(color(black)("mole NaNO"_3)))) = color(darkgreen)(ul(color(black)("260 g NaNO}}_{3}}}}$

The answer must be rounded to two sig figs, the number of sig figs you have for the molarity and volume of the solution.