Question #f2952

The idea here is that in order to find the solution's molarity, you need to find the number of moles of potassium nitrate, the solute, present in exactly #"1 L" = 10^3# #"mL"# of the solution.

To make the calculations easier, start with a sample of #"1 L" = 10^3# #"mL"# of this #22%"m/m"# potassium nitrate solution.

To find the mass of the sample, use the density of the solution.

#10^3 color(red)(cancel(color(black)("mL solution"))) * "1.15 g"/(1color(red)(cancel(color(black)("mL solution")))) = "1150 g"#

Now, you know that this solution is #22%"m/m"# potassium nitrate, which means that every #"100 g"# of this solution will contain #"22 g"# of solute.

This implies that you sample will contain

#1150 color(red)(cancel(color(black)("g solution"))) * "22 g KNO"_3/(100color(red)(cancel(color(black)("g solution")))) = "253 g KNO"_3#

Next, convert the number of grams of potassium nitrate to moles by using the compound's molar mass

#253 color(red)(cancel(color(black)("g"))) * "1 mole KNO"_3/(101.103 color(red)(cancel(color(black)("g")))) = "2.5 moles KNO"_3#

Since this represents the number of moles of potassium present in #"1 L" = 10^3# #"mL"# of this solution, you can say that its molarity will be

#color(darkgreen)(ul(color(black)("molarity = 2.5 mol L"^(-1))))#

The answer is rounded to two sig figs.