# Question #f2952

##### 1 Answer

#### Answer:

#### Explanation:

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

To make the calculations easier, start with a sample of

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 **every**

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 **molarity** will be

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

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