# Question #75537

##### 1 Answer

Here's what I got.

#### Explanation:

Your goal here is to find a way to go from *liters of solution* to **kilograms of solvent** by using the density of the solution.

As you know, a solution's **molarity**,

Let's assume that your starting solution has a molarity of

Let's pick a **moles** of solute in

#V color(red)(cancel(color(black)("L solution"))) * (c color(white)(a)"moles solute")/(1color(red)(cancel(color(black)("L solution")))) = (c * V)color(white)(a)"moles solute"#

Now, the density of the solution is usually given in *grams per milliliter*, which means that you're going to have to convert the volume of the sample from *liters* to *milliliters*

#V color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = (V * 10^3)" mL"#

Use the density of the solution to find its **mass**

#(V * 10^3) color(red)(cancel(color(black)("mL solution"))) * (rhocolor(white)(a)"g")/(1color(red)(cancel(color(black)("mL solution")))) = (rho * V * 10^3)" g"#

You know that this sample contains **moles** of solute. Use the **molar mass** of the solute, let's say *grams* of solute you have in this sample

#(c * V) color(red)(cancel(color(black)("moles solute"))) * (M_Mcolor(white)(a)"g")/(1color(red)(cancel(color(black)("mole solute")))) = (c * V * M_M)" g"#

Since the mass of the sample is made up of the mass of the solute * and* the mass of the solvent, you will have

#m_"solvent" = m_"solution" - m_"solute"#

This will get you

#m_"solvent" = (rho * V * 10^3)" g" - (c * V * M_M)" g" = V * (rho * 10^3 - c * M_M)" g"#

Now, the solution's **molality**, **one kilogram of solvent**. Convert the mass of the solvent from *grams* to *kilograms*

#V * (rho * 10^3 - c * M_M) color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = (V * (rho * 10^3 - c * M_M))/10^3" kg"#

Since the sample contains **moles of solute** in **kilograms of solvent**, its molality will be

#b = (c color(red)(cancel(color(black)(V)))color(white)(a)"moles")/((color(red)(cancel(color(black)(V))) * (rho * 10^3 - c * M_M))/10^3"kg") = color(green)(|bar(ul(color(white)(a/a)color(black)((c * 10^3)/(rho * 10^3 - c * M_M) color(white)(a)"mol kg"^(-1))color(white)(a/a)|)))#

Here

An important thing to notice here is that the molality of the solution is **independent** of the volume of the sample, **the same molality**, and *molarity*, for that matter, regardless of the sample you pick.