Question #324ec

1 Answer
Mar 14, 2017

Answer:

#"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.