# Question b80d8

May 16, 2015

Here's how you can determine the density of a solution if you know its molarity and its molality.

Molarity is defined as moles of solute per liters of solution.

$C = \frac{n}{V} _ \text{solution}$ $\text{ }$$\textcolor{b l u e}{\left(1\right)}$

SIDE NOTE n will always represent the number of moles of solute, so I'll just leave it as n, instead of writing ${n}_{\text{solute}}$.

Since density is defined as mass per unit of volume, you can write the volume of the solution as

$\rho = \frac{m}{V} \implies {V}_{\text{solution" = m_"solution}} / \left(\rho\right)$

Use this in equation $\textcolor{b l u e}{\left(1\right)}$ to get

C = n/(m_"solution"/(rho)) = (n * rho)/m_"solution"# $\text{ } \textcolor{b l u e}{\left(2\right)}$

You can write the mass of the solution as the sum of the mass of the solute and the mass of the solvent

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

Use this in equation $\textcolor{b l u e}{\left(2\right)}$ to get

$C = \frac{n \cdot \rho}{{m}_{\text{solute" + m_"solvent}}}$ $\text{ } \textcolor{b l u e}{\left(3\right)}$

Now use the solution's molality, which is defined as moles of solute per kilograms of solvent.

$b = \frac{n}{m} _ \text{solvent" => m_"solvent} = \frac{n}{b}$

Use this in equation $\textcolor{b l u e}{\left(3\right)}$ to get

$C = \frac{n \cdot \rho}{\left({m}_{\text{solute}} + \frac{n}{b}\right)}$ $\text{ } \textcolor{b l u e}{\left(4\right)}$

Assuming you know what your solute is, you can write its mass as the product between the number of moles and its molar mass

${m}_{\text{solute}} = n \cdot {M}_{M}$

Use this in equation $\textcolor{b l u e}{\left(4\right)}$ to get

$C = \frac{n \cdot \rho}{\left(n \cdot {M}_{M} + \frac{n}{b}\right)} = \frac{\cancel{n} \cdot \rho}{\cancel{n} \left({M}_{M} + \frac{1}{b}\right)} = \frac{\rho}{{M}_{M} + \frac{1}{b}}$ $\text{ } \textcolor{b l u e}{\left(4\right)}$

As a result, the solution's density can be expressed as

$\textcolor{g r e e n}{\rho = C \cdot \left({M}_{M} + \frac{1}{b}\right)}$

Check to see if the units come out right

$\rho = \left(\cancel{\text{moles"))/"L" * ("grams"/cancel("mole") + "kg"/cancel("mole}}\right)$

$\rho = \text{grams"/"L" + "kg"/"L}$

Convert the second term from kg per liter to grams per liter to get

$\rho = \text{grams"/"L" + "grams"/"L" = "grams"/"L}$

And that's how you'd get the density of a solution from its molarity and molality.