Question #67be9

1 Answer
Oct 14, 2016

Answer:

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)|)))#