Question

Question #f2d31

Answer 1

`"0.270 mol L"^(-1)`

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 `"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.