# Question 67be9

Oct 14, 2016

Here's what I got.

#### Explanation:

The first thing to do here is to pick a sample of this $\text{5.0 M}$ potassium chloride solution and use its density to finds its mass.

To make the calculations easier, pick a $\text{1.0-L}$ 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 $\text{5.0-M}$ solution will contain $5.0$ moles of potassium chloride for every liter of solution.

Since we've picked a $\text{1.0-L}$ 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

${m}_{\text{solvent" = m_"solution" - m_"solute}}$

Plug in your values to find

${m}_{\text{solvent" = "1000 g" - "372.75 g" = "627.25 g}}$

So, you know that you get $\text{372.75 g}$ of potassium chloride for every $\text{627.25 g}$ of water, which means that for $\text{100 g}$ 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

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{m}_{\text{KCl" "/100 g H"_2"O" = "59 g}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$