Question

Question #67be9

Answer 1

Here's what I got.

Explanation:

The first thing to do here is to pick a sample of this `"5.0 M"` potassium chloride solution and use its density to finds its mass.

To make the calculations easier, pick a `"1.0-L"` sample. As you know, the molarity of a solution tells you how many moles of solute, which in your case is potassium chloride, you get per liter of solution.

This means that your `"5.0-M"` solution will contain `5.0` moles of potassium chloride for every liter of solution.

Since we've picked a `"1.0-L"` sample, we can say that it will contain `5.0` moles of potassium chloride.

Now, use the density of the solution to find the mass of the sample

`1.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.0 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("the given density")) = "1000 g"`

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

`5.0 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "372.75 g"`

Now, the mass of the solvent will be equal to the difference between the mass of the solution and the mass of the solute

`m_"solvent" = m_"solution" - m_"solute"`

Plug in your values to find

`m_"solvent" = "1000 g" - "372.75 g" = "627.25 g"`

So, you know that you get `"372.75 g"` of potassium chloride for every `"627.25 g"` of water, which means that for `"100 g"` of water you will have

`100 color(red)(cancel(color(black)("g solvent"))) * "372.75 g KCl"/(627.25color(red)(cancel(color(black)("g solvent")))) = "59.43 g KCl"`

Rounded to two sig figs, the answer will be

`color(green)(bar(ul(|color(white)(a/a)color(black)(m_ "KCl" "/100 g H"_2"O" = "59 g")color(white)(a/a)|)))`