# Question #f2d31

##### 1 Answer

#### Explanation:

Your starting point here will be to pick a sample of this potassium chloride solution. To make the calculations easier, let's pick a sample that contains exactly

As you know, the **molality** of the solution tells you the number of moles of solute present in exactly

In your case, a **moles** of potassium chloride, the solute, for every **moles** of potassium chloride.

Use the **molar mass** of potassium chloride to convert the number of moles to *grams*

#0.273 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "20.35 g"#

So, this sample contains **total mass** of

#10^3color(white)(.)"g" + "20.35 g" = "1020.35 g"#

Next, use the **density** of the solution to find its volume.

#1020.35 color(red)(cancel(color(black)("g solution"))) * "1 L solution"/(1.011 * 10^3color(red)(cancel(color(black)("g")))) = "1.00925 L solution"#

Now, in order to find the **molarity** of the solution, you need to figure out how many moles of solute are present in exactly

Use the fact that **moles** of potassium chloride to find the number of moles present in exactly #"1 L"3 of the solution.

#1 color(red)(cancel(color(black)("L solution"))) * "0.273 moles KCl"/(1.00925color(red)(cancel(color(black)("L solution")))) = "0.270 moles KCl"#

You can thus say that this solution has a molarity of

#color(darkgreen)(ul(color(black)("molarity = 0.270 mol L"^(-1))))#

The answer is rounded to three **sig figs**, the number of sifg figs you have for your data.