Question #b80d8

1 Answer
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 = n/V_"solution" " " color(blue)((1))

SIDE NOTE n will always represent the number of moles of solute, so I'll just leave it as n, instead of writing n_"solute".

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

rho = m/V => V_"solution" = m_"solution"/(rho)

Use this in equation color(blue)((1)) to get

C = n/(m_"solution"/(rho)) = (n * rho)/m_"solution" " "color(blue)((2))

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

m_"solution" = m_"solute" + m_"solvent"

Use this in equation color(blue)((2)) to get

C = (n * rho)/(m_"solute" + m_"solvent") " "color(blue)((3))

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

b = n/m_"solvent" => m_"solvent" = n/b

Use this in equation color(blue)((3)) to get

C = (n * rho)/((m_"solute" + n/b)) " "color(blue)((4))

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_"solute" = n * M_M

Use this in equation color(blue)((4)) to get

C = (n * rho)/((n * M_M + n/b)) = (cancel(n) * rho)/(cancel(n)(M_M + 1/b)) = (rho)/(M_M + 1/b) " "color(blue)((4))

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

color(green)(rho = C * (M_M + 1/b))

Check to see if the units come out right

rho = (cancel("moles"))/"L" * ("grams"/cancel("mole") + "kg"/cancel("mole"))

rho = "grams"/"L" + "kg"/"L"

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

rho = "grams"/"L" + "grams"/"L" = "grams"/"L"

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