# Question #67be9

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The first thing to do here is to pick a sample of this **density** to finds its **mass**.

To make the calculations easier, pick a **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 **moles** of potassium chloride *for every liter of solution*.

Since we've picked a **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 **for every** **water**, which means that for

#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)|)))#