# Calculate the molality of a solution made by dissolving #"115 g"# of #"NaNO"_3# in #"500. mL"# of water? The molar mass of #"NaNO"_3# is #"85.0 g/mol"# and the density of water is #"1.00 g/mL"#.

##### 1 Answer

#### Answer:

#### Explanation:

As you know, the **molality** of a solution tells you the number of moles of solute present for every **solvent**.

This means that the first thing that you need to do here is to figure out how many *grams* of water are present in your sample. To do that, use the **density** of water.

#500. color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "500. g"#

Next, use the **molar mass** of the solute to determine how many *moles* are present in the sample.

#115 color(red)(cancel(color(black)("g"))) * "1 mole NanO"_3/(85.0color(red)(cancel(color(black)("g")))) = "1.353 moles NaNO"_3#

So, you know that this solution will contain **moles** of sodium nitrate, the solute, for

In order to find the **molality** of the solution, you must figure out how many moles of solute would be present for

#10^3 color(red)(cancel(color(black)("g water"))) * "1.353 moles NaNO"_3/(500. color(red)(cancel(color(black)("g water")))) = "2.706 moles NaNO"_3#

You can thus say that the molality of the solution is equal to

#color(darkgreen)(ul(color(black)("molality"))) = "2.706 mol kg"^(-1) ~~ color(darkgreen)(ul(color(black)("2.71 mol kg"^(-1)))#

The answer is rounded to three **sig figs**.

So, you know that this solution will contain **moles** of solute for every