Question

Question #324ec

Answer 1

`"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 `"1 L"` of solution.

In this case, a `"1.5-M"` sodium nitrate solution will contain `1.5` moles of sodium nitrate, the solute, for every `"1 L"` of solution.

In other words, regardless if you scale this solution up or down, i.e. if you make more than `"1 L"` of less than `"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 `"2 L"`, so it must contain twice as many moles as `"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 color(red)(cancel(color(black)("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.