# Question 75537

Jul 4, 2016

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, $c$, tells you how many moles of solute you have in one liter of solution.

Let's assume that your starting solution has a molarity of $c$ $\text{M}$ and a density of $\rho$ ${\text{g mL}}^{- 1}$. The first thing to do here is pick a sample of this initial solution.

Let's pick a $V$ $\text{L}$ sample. A solution that has a molarity of $c$ $\text{M}$ will contain $c$ moles of solute in $\text{1 L}$ of solution, which means that $V$ $\text{L}$ will contain

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 $\left(c \cdot V\right)$ moles of solute. Use the molar mass of the solute, let's say ${M}_{M}$ ${\text{g mol}}^{- 1}$, to find how many 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

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

This will get you

${m}_{\text{solvent" = (rho * V * 10^3)" g" - (c * V * M_M)" g" = V * (rho * 10^3 - c * M_M)" g}}$

Now, the solution's molality, $b$, is calculated by taking the number of moles of solute present in 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 $\left(c \cdot V\right)$ moles of solute in $\frac{V \cdot \left(\rho \cdot {10}^{3} - c \cdot {M}_{M}\right)}{10} ^ 3$ 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

$c$ - the molarity of the solution in ${\text{mol L}}^{- 1}$
$\rho$ - the density of the solution in ${\text{g mol}}^{- 1}$
${M}_{M}$ - the molar mass of the solute in ${\text{g mol}}^{- 1}$

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