Question #f2d31

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 #"1 kg"# of water, the solvent.

As you know, the molality of the solution tells you the number of moles of solute present in exactly #"1 kg"# of the solvent.

In your case, a #"0.273-m"# potassium chloride solution contains #0.273# moles of potassium chloride, the solute, for every #"1 kg"# of water. This implies that our sample, which contains exactly #"1 kg"# of water, will also contain #0.273# 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 #"1 kg" = 10^3# #"g"# of water and #"20.35 g"# of potassium chloride, which means that it has a 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 #"1 L"# of the solution.

Use the fact that #"1.00925 L"# of this solution contain #0.273# 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.